Calculating the equation of a curve - GMAT Quantitative

Card 0 of 72

Question

Suppose the points and are plotted to connect a line. What are the -intercept and -intercept, respectively?

Answer

First, given the two points, find the equation of the line using the slope formula and the y-intercept equation.

Slope:

$\frac{y_2-y_1}{x_2-x_1} = \frac{3-(-2)}{1-(-3)}=\frac{5}{4}$

Write the slope-intercept formula.

Substitute a given point and the slope into the equation to find the y-intercept.

$3=\left(\frac{5}{4}\right)(1)+b$

$\frac{12}{4}=\frac{5}{4}+b$

$b=\frac{7}{4}$

The y-intercept is: $b=\frac{7}{4}$ .

Substitiute the slope and the y-intercept into the slope-intercept form.

$y=\frac{5}{4}x+\frac{7}{4}$

To find the x-intercept, substitute and solve for x.

$0=\frac{5}{4}x+\frac{7}{4}$

$-\frac{7}{4}=\frac{5}{4}x$

$-\frac{7}{4} \cdot (\frac{4}{5})=\frac{5}{4}x \cdot (\frac{4}{5})$

$x=-\frac{7}{5}$

The x-intercept is: $x=-\frac{7}{5}$

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Question

Suppose the curve of a function is parabolic. The -intercept is and the vertex is the -intercept at . What is a possible equation of the parabola, if it exists?

Answer

Write the standard form of the parabola.

Given the point , the y-intercept is -4, which indicates that . This is also the vertex, so the vertex formula can allow writing an expression in terms of variables and .

Write the vertex formula and substitute the known vertex given point .

$x=-\frac{b}{2a}$

$0=-\frac{b}{2a}$

Using the values of , , and the other given point , substitute these values to the standard form and solve for .

Substitute the values of ,, and into the standard form of the parabola.

The correct answer is:

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Question

If the -intercept and the slope are , what's the equation of the line in standard form?

Answer

Write the slope intercept formula.

Convert the given x-intercept to a known point, which is .

Substitute the given slope and the point to solve for the y-intercept.

Substitute the slope and y-intercept into the slope-intercept formula.

Add 1 on both sides of the equation, and subtract on both sides of the equation to find the equation in standard form.

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Question

Which of the following functions has as its graph a curve with , and as its only two -intercepts?

Answer

By the Fundamental Theorem of Algebra, a polynomial equation of degree 3 must have three solutions, or roots, but one root can be a double root or triple root. Since the polynomial here has two roots, and 4, one of these must be a double root. Since the leading term is $x^{2}$ , the equation must be

or

We rewrite both.

$(x+4)(x^{2}-4^{2}) = 0$

$(x+4)(x^{2}-16) = 0$

$x \cdot x^{2}- x \cdot 16 + 4 \cdot x^{2}- 4 \cdot 16 = 0$

$x^{3}-16 x + 4 x^{2}- 64 = 0$

$x^{3}+ 4 x^{2}-16 x - 64 = 0$

$(x^{2}-16) (x-4)= 0$

$x \cdot x^{2}- 4 \cdot x^{2}- x \cdot 16 + 4 \cdot 16 = 0$

$x^{3}- 4 x^{2}-16 x + 64 = 0$

The correct response can be $f(x) = x^{3}+ 4 x^{2}-16 x - 64$ or $f(x) = x^{3}- 4 x^{2}-16 x + 64$ . The first is not among the choices, so the last is the correct choice.

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Question

Which of the following functions has as its graph a curve with -intercepts , , and ?

Answer

A polynomial equation of degree 3 with solution set $\left { -2, 2, 3\right }$ and leading term $x^{2}$ takes the form

We can rewrite this as follows:

$(x^{2}-2^{2})(x-3)= 0$

$(x^{2}-4)(x-3)= 0$

$x^{2}\cdot x- x^{2} \cdot 3 - 4\cdot x + 4 \cdot 3 = 0$

$x^{3} - 3 x^{2} - 4 x + 12 = 0$

The correct response is $f(x)= x^{3} - 3 x^{2} - 4 x + 12$ .

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Question

Which of the following functions does not have as its graph a curve with as an -intercept?

Answer

We can evaluate in each of the definitions of in the five choices. If , is an -intercept.

$f(x) = x^{3}+ 5x^{2} -25 x-125$

$f(-5) = (-5)^{3}+ 5 \cdot (-5)^{2} -25 (-5)-125$

$= (-5)^{3}+ 5 \cdot 25 - (-125)-125$

$f(x) = x^{3}- 5x^{2} -25 x+125$

$f(-5) = (-5)^{3}- 5 \cdot (-5)^{2} -25 (-5)+125$

$= (-5)^{3}+ 5 \cdot 25 - (-125)-125$

$f(x) = x^{3} +125$

$f(-5) = (-5) ^{3} +125$

$f(x) = x^{3}+1 5x^{2} + 75 x+125$

$f(-5) = (-5)^{3}+1 5\cdot (-5)^{2} + 75 (-5)+125$

$f(-5) = -125+1 5\cdot 25 + (-375)+125$

$f(x) = x^{3}-1 5x^{2} + 75 x-125$

$f(-5) = (-5)^{3}-1 5\cdot (-5)^{2} + 75 (-5)-125$

$f(-5) = -125-1 5\cdot 25 + (-375)-125$

$f(x) = x^{3}-1 5x^{2} + 75 x-125$ does not have as an -intercept, so it is the correct choice.

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Question

Only one of the following equations has a graph with an -intercept between $\left ( \frac{1}{2},0 \right )$ and . Which one?

Answer

The Intermediate Value Theorem states that if is a continuous function, as all five of the polynomial functions in the given choices are, and $f\left ( \frac{1}{2} \right )$ and are of different sign, then the graph of has an -intercept on the interval $\left ( \frac{1}{2}, 1 \right )$ .

We evaluate $f\left ( \frac{1}{2} \right )$ and for each of the five choices to find the one for which the two have different sign.

$f(x) = x^{4}-x^{3}+ x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+ \frac{1}{2} -3 = \frac{1}{16} - \frac{1}{8}+ \frac{1}{2}-3=-2\frac{9}{16}$

$f(1) = 1^{4}-1^{3}+ 1-3 = 1-1+1-3 = -2$

$f\left ( \frac{1}{2} \right )$ and are both negative.

$f(x) = x^{4}-x^{3}+ 2 x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+2 \left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+1-3=-2\frac{1}{16}$

$f(1) = 1^{4}-1^{3}+ 2 \cdot 1-3 = 1-1+2-3 = -1$

$f\left ( \frac{1}{2} \right )$ and are both negative.

$f(x) = x^{4}-x^{3}+ 4 x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+4 \left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+2-3=-1\frac{1}{16}$

$f(1) = 1^{4}-1^{3}+ 4 \cdot 1-3 = 1-1+4-3 = 1$

$f\left ( \frac{1}{2} \right )$ and are of different sign.

$f(x) = x^{4}-x^{3}+ 8 x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+8\left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+4-3= \frac{15}{16}$

$f(1) = 1^{4}-1^{3}+ 8 \cdot 1-3 = 1-1+8-3 = 5$

$f\left ( \frac{1}{2} \right )$ and are both positive.

$f(x) = x^{4}-x^{3}+ 16x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+16\left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+8-3=4 \frac{15}{16}$

$f(1) = 1^{4}-1^{3}+ 16 \cdot 1-3 = 1-1+16-3 = 13$

$f\left ( \frac{1}{2} \right )$ and are both positive.

$f(x) = x^{4}-x^{3}+ 4 x-3$ is the function in which $f\left ( \frac{1}{2} \right )$ and are of different sign, so it is represented by a graph with an -intercept between $\left ( \frac{1}{2},0 \right )$ and . This is the correct choice.

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Question

Between which two points is an -intercept of the graph of the function

$f(x) = x^{4}+3x^{2}-17x-115$

located?

Answer

As a polynomial function, has a continuous graph. By the Intermediate Value Theorem, if and are of different sign, then for some $c \in (a,b)$ - that is, the graph of has an -intercept between and . Evaluate for all and observe between which two integers the sign changes.

$f(x) = x^{4}+3x^{2}-17x-115$

$f(0) = 0^{4}+3 \cdot 0^{2}-17\cdot 0-115$

$f(1) =1^{4}+3\cdot 1^{2}-17 \cdot 1-115$

$f(2) = 2^{4}+3\cdot 2^{2}-17\cdot 2-115$

$= 16+3\cdot 4 -34 -115$

$f(3) =3 ^{4}+3 \cdot 3^{2}-17 \cdot 3-115$

$= 81 + 3 \cdot 9 - 51 -115$

$f(4) = 4 ^{4}+3\cdot 4^{2}-17\cdot 4 -115$

$= 256 +3\cdot 16 -68-115$

$f(5) = 5^{4}+3 \cdot 5^{2}-17 \cdot 5 -115$

$= 625 +3 \cdot 25 -85 -115$

Since and , the -intercept is between and .

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Question

A function is defined as

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

where are integer coefficients whose values (which might be positive, negative, or zero) are not given. Which of the following cannot be an -intercept of the graph of no matter what the values of those three coefficients are?

Answer

Since the graph of a function has its -intercept at a point if and only if , finding possible -intercepts of the graph of is equivalent to finding a solution of . Since has integer coefficients, then by the Rational Zeroes Theorem, any rational solutions to the equation

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of $\left { 1, 2, 3, 4, 6, 12 \right }$ - and a factor of leading coefficient 2 - that is, an element of $\left { 1,2 \right }$ . Since all of the choices are positive, we will only look at possible positive solutions.

The quotients of an element of the first set and an element of the last are:

$\frac{1}{1} = 1$ ; $\frac{2}{1} = 2$ ; $\frac{3}{1} = 3$ ; $\frac{4}{1} = 4$ ; $\frac{6}{1} = 6$ ; $\frac{12}{1} = 12$ ;

$\frac{1}{2}$ ; $\frac{2}{2} = 1$ ; $\frac{3}{2}$ ; $\frac{4}{2} = 2$ ; $\frac{6}{2} = 3$ ; $\frac{12}{2} = 6$

Eliminating duplicates, the set of possible positive rational solutions to is

$\left { \frac{1}{2},1, \frac{3}{2} , 2,3,4,6,12 \right }$ .

Of the five choices, only $\frac{1}{6}$ does not appear in the set of possible rational solutions of , so of the five choices, only $\left ( \frac{1}{6}, 0 \right )$ cannot be an -intercept of the graph.

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Question

Suppose the points and are plotted to connect a line. What are the -intercept and -intercept, respectively?

Answer

First, given the two points, find the equation of the line using the slope formula and the y-intercept equation.

Slope:

$\frac{y_2-y_1}{x_2-x_1} = \frac{3-(-2)}{1-(-3)}=\frac{5}{4}$

Write the slope-intercept formula.

Substitute a given point and the slope into the equation to find the y-intercept.

$3=\left(\frac{5}{4}\right)(1)+b$

$\frac{12}{4}=\frac{5}{4}+b$

$b=\frac{7}{4}$

The y-intercept is: $b=\frac{7}{4}$ .

Substitiute the slope and the y-intercept into the slope-intercept form.

$y=\frac{5}{4}x+\frac{7}{4}$

To find the x-intercept, substitute and solve for x.

$0=\frac{5}{4}x+\frac{7}{4}$

$-\frac{7}{4}=\frac{5}{4}x$

$-\frac{7}{4} \cdot (\frac{4}{5})=\frac{5}{4}x \cdot (\frac{4}{5})$

$x=-\frac{7}{5}$

The x-intercept is: $x=-\frac{7}{5}$

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Question

Suppose the curve of a function is parabolic. The -intercept is and the vertex is the -intercept at . What is a possible equation of the parabola, if it exists?

Answer

Write the standard form of the parabola.

Given the point , the y-intercept is -4, which indicates that . This is also the vertex, so the vertex formula can allow writing an expression in terms of variables and .

Write the vertex formula and substitute the known vertex given point .

$x=-\frac{b}{2a}$

$0=-\frac{b}{2a}$

Using the values of , , and the other given point , substitute these values to the standard form and solve for .

Substitute the values of ,, and into the standard form of the parabola.

The correct answer is:

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Question

If the -intercept and the slope are , what's the equation of the line in standard form?

Answer

Write the slope intercept formula.

Convert the given x-intercept to a known point, which is .

Substitute the given slope and the point to solve for the y-intercept.

Substitute the slope and y-intercept into the slope-intercept formula.

Add 1 on both sides of the equation, and subtract on both sides of the equation to find the equation in standard form.

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Question

Which of the following functions has as its graph a curve with , and as its only two -intercepts?

Answer

By the Fundamental Theorem of Algebra, a polynomial equation of degree 3 must have three solutions, or roots, but one root can be a double root or triple root. Since the polynomial here has two roots, and 4, one of these must be a double root. Since the leading term is $x^{2}$ , the equation must be

or

We rewrite both.

$(x+4)(x^{2}-4^{2}) = 0$

$(x+4)(x^{2}-16) = 0$

$x \cdot x^{2}- x \cdot 16 + 4 \cdot x^{2}- 4 \cdot 16 = 0$

$x^{3}-16 x + 4 x^{2}- 64 = 0$

$x^{3}+ 4 x^{2}-16 x - 64 = 0$

$(x^{2}-16) (x-4)= 0$

$x \cdot x^{2}- 4 \cdot x^{2}- x \cdot 16 + 4 \cdot 16 = 0$

$x^{3}- 4 x^{2}-16 x + 64 = 0$

The correct response can be $f(x) = x^{3}+ 4 x^{2}-16 x - 64$ or $f(x) = x^{3}- 4 x^{2}-16 x + 64$ . The first is not among the choices, so the last is the correct choice.

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Question

Which of the following functions has as its graph a curve with -intercepts , , and ?

Answer

A polynomial equation of degree 3 with solution set $\left { -2, 2, 3\right }$ and leading term $x^{2}$ takes the form

We can rewrite this as follows:

$(x^{2}-2^{2})(x-3)= 0$

$(x^{2}-4)(x-3)= 0$

$x^{2}\cdot x- x^{2} \cdot 3 - 4\cdot x + 4 \cdot 3 = 0$

$x^{3} - 3 x^{2} - 4 x + 12 = 0$

The correct response is $f(x)= x^{3} - 3 x^{2} - 4 x + 12$ .

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Question

Which of the following functions does not have as its graph a curve with as an -intercept?

Answer

We can evaluate in each of the definitions of in the five choices. If , is an -intercept.

$f(x) = x^{3}+ 5x^{2} -25 x-125$

$f(-5) = (-5)^{3}+ 5 \cdot (-5)^{2} -25 (-5)-125$

$= (-5)^{3}+ 5 \cdot 25 - (-125)-125$

$f(x) = x^{3}- 5x^{2} -25 x+125$

$f(-5) = (-5)^{3}- 5 \cdot (-5)^{2} -25 (-5)+125$

$= (-5)^{3}+ 5 \cdot 25 - (-125)-125$

$f(x) = x^{3} +125$

$f(-5) = (-5) ^{3} +125$

$f(x) = x^{3}+1 5x^{2} + 75 x+125$

$f(-5) = (-5)^{3}+1 5\cdot (-5)^{2} + 75 (-5)+125$

$f(-5) = -125+1 5\cdot 25 + (-375)+125$

$f(x) = x^{3}-1 5x^{2} + 75 x-125$

$f(-5) = (-5)^{3}-1 5\cdot (-5)^{2} + 75 (-5)-125$

$f(-5) = -125-1 5\cdot 25 + (-375)-125$

$f(x) = x^{3}-1 5x^{2} + 75 x-125$ does not have as an -intercept, so it is the correct choice.

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Question

Only one of the following equations has a graph with an -intercept between $\left ( \frac{1}{2},0 \right )$ and . Which one?

Answer

The Intermediate Value Theorem states that if is a continuous function, as all five of the polynomial functions in the given choices are, and $f\left ( \frac{1}{2} \right )$ and are of different sign, then the graph of has an -intercept on the interval $\left ( \frac{1}{2}, 1 \right )$ .

We evaluate $f\left ( \frac{1}{2} \right )$ and for each of the five choices to find the one for which the two have different sign.

$f(x) = x^{4}-x^{3}+ x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+ \frac{1}{2} -3 = \frac{1}{16} - \frac{1}{8}+ \frac{1}{2}-3=-2\frac{9}{16}$

$f(1) = 1^{4}-1^{3}+ 1-3 = 1-1+1-3 = -2$

$f\left ( \frac{1}{2} \right )$ and are both negative.

$f(x) = x^{4}-x^{3}+ 2 x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+2 \left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+1-3=-2\frac{1}{16}$

$f(1) = 1^{4}-1^{3}+ 2 \cdot 1-3 = 1-1+2-3 = -1$

$f\left ( \frac{1}{2} \right )$ and are both negative.

$f(x) = x^{4}-x^{3}+ 4 x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+4 \left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+2-3=-1\frac{1}{16}$

$f(1) = 1^{4}-1^{3}+ 4 \cdot 1-3 = 1-1+4-3 = 1$

$f\left ( \frac{1}{2} \right )$ and are of different sign.

$f(x) = x^{4}-x^{3}+ 8 x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+8\left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+4-3= \frac{15}{16}$

$f(1) = 1^{4}-1^{3}+ 8 \cdot 1-3 = 1-1+8-3 = 5$

$f\left ( \frac{1}{2} \right )$ and are both positive.

$f(x) = x^{4}-x^{3}+ 16x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+16\left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+8-3=4 \frac{15}{16}$

$f(1) = 1^{4}-1^{3}+ 16 \cdot 1-3 = 1-1+16-3 = 13$

$f\left ( \frac{1}{2} \right )$ and are both positive.

$f(x) = x^{4}-x^{3}+ 4 x-3$ is the function in which $f\left ( \frac{1}{2} \right )$ and are of different sign, so it is represented by a graph with an -intercept between $\left ( \frac{1}{2},0 \right )$ and . This is the correct choice.

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Question

Between which two points is an -intercept of the graph of the function

$f(x) = x^{4}+3x^{2}-17x-115$

located?

Answer

As a polynomial function, has a continuous graph. By the Intermediate Value Theorem, if and are of different sign, then for some $c \in (a,b)$ - that is, the graph of has an -intercept between and . Evaluate for all and observe between which two integers the sign changes.

$f(x) = x^{4}+3x^{2}-17x-115$

$f(0) = 0^{4}+3 \cdot 0^{2}-17\cdot 0-115$

$f(1) =1^{4}+3\cdot 1^{2}-17 \cdot 1-115$

$f(2) = 2^{4}+3\cdot 2^{2}-17\cdot 2-115$

$= 16+3\cdot 4 -34 -115$

$f(3) =3 ^{4}+3 \cdot 3^{2}-17 \cdot 3-115$

$= 81 + 3 \cdot 9 - 51 -115$

$f(4) = 4 ^{4}+3\cdot 4^{2}-17\cdot 4 -115$

$= 256 +3\cdot 16 -68-115$

$f(5) = 5^{4}+3 \cdot 5^{2}-17 \cdot 5 -115$

$= 625 +3 \cdot 25 -85 -115$

Since and , the -intercept is between and .

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Question

A function is defined as

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

where are integer coefficients whose values (which might be positive, negative, or zero) are not given. Which of the following cannot be an -intercept of the graph of no matter what the values of those three coefficients are?

Answer

Since the graph of a function has its -intercept at a point if and only if , finding possible -intercepts of the graph of is equivalent to finding a solution of . Since has integer coefficients, then by the Rational Zeroes Theorem, any rational solutions to the equation

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of $\left { 1, 2, 3, 4, 6, 12 \right }$ - and a factor of leading coefficient 2 - that is, an element of $\left { 1,2 \right }$ . Since all of the choices are positive, we will only look at possible positive solutions.

The quotients of an element of the first set and an element of the last are:

$\frac{1}{1} = 1$ ; $\frac{2}{1} = 2$ ; $\frac{3}{1} = 3$ ; $\frac{4}{1} = 4$ ; $\frac{6}{1} = 6$ ; $\frac{12}{1} = 12$ ;

$\frac{1}{2}$ ; $\frac{2}{2} = 1$ ; $\frac{3}{2}$ ; $\frac{4}{2} = 2$ ; $\frac{6}{2} = 3$ ; $\frac{12}{2} = 6$

Eliminating duplicates, the set of possible positive rational solutions to is

$\left { \frac{1}{2},1, \frac{3}{2} , 2,3,4,6,12 \right }$ .

Of the five choices, only $\frac{1}{6}$ does not appear in the set of possible rational solutions of , so of the five choices, only $\left ( \frac{1}{6}, 0 \right )$ cannot be an -intercept of the graph.

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Question

Suppose the points and are plotted to connect a line. What are the -intercept and -intercept, respectively?

Answer

First, given the two points, find the equation of the line using the slope formula and the y-intercept equation.

Slope:

$\frac{y_2-y_1}{x_2-x_1} = \frac{3-(-2)}{1-(-3)}=\frac{5}{4}$

Write the slope-intercept formula.

Substitute a given point and the slope into the equation to find the y-intercept.

$3=\left(\frac{5}{4}\right)(1)+b$

$\frac{12}{4}=\frac{5}{4}+b$

$b=\frac{7}{4}$

The y-intercept is: $b=\frac{7}{4}$ .

Substitiute the slope and the y-intercept into the slope-intercept form.

$y=\frac{5}{4}x+\frac{7}{4}$

To find the x-intercept, substitute and solve for x.

$0=\frac{5}{4}x+\frac{7}{4}$

$-\frac{7}{4}=\frac{5}{4}x$

$-\frac{7}{4} \cdot (\frac{4}{5})=\frac{5}{4}x \cdot (\frac{4}{5})$

$x=-\frac{7}{5}$

The x-intercept is: $x=-\frac{7}{5}$

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Question

Suppose the curve of a function is parabolic. The -intercept is and the vertex is the -intercept at . What is a possible equation of the parabola, if it exists?

Answer

Write the standard form of the parabola.

Given the point , the y-intercept is -4, which indicates that . This is also the vertex, so the vertex formula can allow writing an expression in terms of variables and .

Write the vertex formula and substitute the known vertex given point .

$x=-\frac{b}{2a}$

$0=-\frac{b}{2a}$

Using the values of , , and the other given point , substitute these values to the standard form and solve for .

Substitute the values of ,, and into the standard form of the parabola.

The correct answer is:

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