How to graph a quadratic function - Geometry

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Question

$Find\ the\ vertex\ for\ the\ function: f(x)=4x^{2}-16x+11.$

Answer

$Use \frac{-b}{2a}\ to\ calculate\ the\ x-value\ of\ the\ vertex.$

$\small Given\ the\ function\ f(x)=4x^{2}-16x+11\ is\ in\ standard\ form\ y=ax^{2}+bx+c$

$\small We\ can\ see\ that\ a=4, b=-16\ and\ c=11$

$\small Substitute\ these\ values\ in\ for\ x=\frac{-b}{2a}\ to\ get\ x=\frac{16}{8}=2$

$\small Now\ substitute\ x=2\ into\ the\ function\ f(x)\ to\ get:$

$\small f(2)=4(2)^{2}-16(2)+11=-5$

$\small The\ coordinates\ of\ the\ vertex\ are\ (2,-5)$

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Question

Which of the following is the equation of the line of symmetry of a vertical parabola on the coordinate plane with its vertex at ?

Answer

The line of symmetry of a vertical parabola is a vertical line, the equation of which takes the form for some . The line of symmetry passes through the vertex, which here is , so the equation must be .

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Question

What are the possible values of if the parabola of the quadratic function $f (x) = Ax^{2} + 4x- 2$ is concave upward and does not intersect the -axis?

Answer

If the graph of $f (x) = Ax^{2} + 4x- 2$ is concave upward, then .

If the graph does not intersect the -axis, then $Ax^{2} + 4x- 2 = 0$ has no real solution, and the discriminant $4^{2} - 4 \cdot A \cdot \left ( -2\right )$ is negative:

$4^{2} - 4 \cdot A \cdot \left ( -2\right ) < 0$

$8 A \div 8< - 16 \div 8$

For the parabola to have both characteristics, it must be true that and , but these two events are mutually exclusive. Therefore, the parabola cannot exist.

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Question

Which of the following is the equation of the line of symmetry of a horizontal parabola on the coordinate plane with its vertex at ?

Answer

The line of symmetry of a horizontal parabola is a horizontal line, the equation of which takes the form for some . The line of symmetry passes through the vertex, which here is , so the equation must be .

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Question

A vertical parabola has two -intercepts, one at and one at .

Which of the following must be true about this parabola?

Answer

A parabola with its -intercepts at and at has as its equation

for some nonzero . If this is multiplied out, the equation can be rewritten as

$y = a(x^{2}-12x+20)$

or, simplified,

$y = a x^{2}-12ax+20a$

The sign of quadratic coefficient determines whether it is concave upward or concave downward. We do not have the sign or any way of determining it.

The -coordinate of the -intercept is the contant, , but without knowing , we have no way of knowing .

The -coordinate of the vertex of $y = a x^{2}+bx+c$ is the value $- \frac{b}{2a}$ . since , this expression becomes

$- \frac{b}{2a} = - \frac{-12a}{2a} = 6$

The -coordinate is

$y = a \cdot 6^{2}-12a\cdot 6+20a = 36a - 72a+20a = -16a$ ,

but without knowing , this coordinate, and the vertex itself, cannot be determined.

The line of symmetry is the line $x = - \frac{b}{2a}$ ; this value was computed to be equal to 6, so the line can be determined to be .

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Question

Give the vertex of the graph of the function

$f(x) = 2x^{2}- 8x + 17$ .

Answer

This can be answered rewriting this expression in the form

$f(x) = a(x-h)^{2}+ k$ .

Once this is done, we can identify the vertex as the point .

$f(x) = 2x^{2}- 8x + 17$

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

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

$f(x) = 2(x-2) ^{2}- 8+ 17$

$f(x) = 2(x-2) ^{2}+9$

The vertex is

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Question

$f(x)= Ax^{2}+ Bx+ C$ has as its graph a vertical parabola on the coordinate plane. You are given that , but you are given neither nor .

Which of the following can you determine without knowing the values of and ?

I) Whether the curve opens upward or opens downward

II) The location of the vertex

III) The location of the -intercept

IV) The locations of the -intercepts, if there are any

V) The equation of the line of symmetry

Answer

I) The orientation of the parabola is determined solely by the value of . Since , the parabola can be determined to open upward.

II and V) The -coordinate of the vertex is $-\frac{B}{2A}$ ; since you are not given , you cannot find this. Also, since the line of symmetry has equation $x=-\frac{B}{2A}$ , for the same reason, you cannot find this either.

III) The -intercept is the point at which ; by substitution, it can be found to be at . is unknown, so the -intercept cannot be found.

IV) The -intercept(s), if any, are the point(s) at which $f(x)= Ax^{2}+ Bx+ C = 0$ . This is solvable using the quadratic formula

$x= \frac{-B \pm \sqrt{B^{2}-4AC}}{2A}$

Since all three of and must be known for this to be evaluated, and only is known, the -intercept(s) cannot be identified.

The correct response is I only.

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Question

$f(x)= Ax^{2}+ Bx+ C$ has as its graph a vertical parabola on the coordinate plane. You are given that and , but you are not given .

Which of the following can you determine without knowing the value of ?

I) Whether the graph is concave upward or concave downward

II) The location of the vertex

III) The location of the -intercept

IV) The locations of the -intercepts, if there are any

V) The equation of the line of symmetry

Answer

I) The orientation of the parabola is determined solely by the sign of . Since , the parabola can be determined to be concave downward.

II and V) The -coordinate of the vertex is $-\frac{B}{2A}$ ; since you are not given , you cannot find this. Also, since the line of symmetry has equation $x=-\frac{B}{2A}$ , for the same reason, you cannot find this either.

III) The -intercept is the point at which ; by substitution, it can be found to be at . known to be equal to 9, so the -intercept can be determined to be .

IV) The -intercept(s), if any, are the point(s) at which $f(x)= Ax^{2}+ Bx+ C = 0$ . This is solvable using the quadratic formula

$x= \frac{-B \pm \sqrt{B^{2}-4AC}}{2A}$

Since all three of and must be known for this to be evaluated, and only is known, the -intercept(s) cannot be identified.

The correct response is I and III only.

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Question

$f(x)= Ax^{2}+ Bx+ C$ has as its graph a vertical parabola on the coordinate plane. You are given that and , but you are not given the value of .

Which of the following can you determine without knowing the value of ?

I) Whether the graph is concave upward or concave downward

II) The location of the vertex

III) The location of the -intercept

IV) The locations of the -intercepts, if there are any

V) The equation of the line of symmetry

Answer

I) The orientation of the parabola is determined solely by the sign of . Since , a positive value, the parabola can be determined to be concave upward.

II) The -coordinate of the vertex is $-\frac{B}{2A}$ ; since you given both and , you can find this to be

$h = -\frac{B}{2A} = -\frac{10}{2 (4)} =- \frac{5}{4}$

The -coordinate is equal to $f\left ( \frac{-5}{4} \right )$ . However, you need the entire equation to determine this value; since you do not know , you cannot find the -coordinate. Therefore, you cannot find the vertex.

III) The -intercept is the point at which ; by substitution, it can be found to be at . is unknown, so the -intercept cannot be found.

IV) The -intercept(s), if any, are the point(s) at which $f(x)= Ax^{2}+ Bx+ C = 0$ . This is solvable using the quadratic formula

$x= \frac{-B \pm \sqrt{B^{2}-4AC}}{2A}$

Since all three of and must be known for this to be evaluated, and is unknown, the -intercept(s) cannot be identified.

V) The line of symmetry has equation $x=-\frac{B}{2A}$ . When exploring the vertex, we found that this value is equal to $- \frac{5}{4}$ , so the line of symmetry is the line of the equation $x=- \frac{5}{4}$ .

The correct response is I and V only.

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Question

$f(x)= Ax^{2}+ Bx+ C$ has as its graph a vertical parabola on the coordinate plane. You are given that , but you are given no other information about these values.

Which of the following can you determine without knowing the value of ?

I) Whether the graph is concave upward or concave downward

II) The location of the vertex

III) The location of the -intercept

IV) The locations of the -intercepts,or whether there are any

V) The equation of the line of symmetry

Answer

I) The orientation of the parabola is determined solely by the sign of . It is given in the problem that is negative, so it follows that the parabola is concave downward.

II) The -coordinate of the vertex is $-\frac{B}{2A}$ ; since , this number is $-\frac{A}{2A} = -\frac{1}{2}$ . The -coordinate is $f\left (- \frac{1}{2} \right )$ , but since we do not know the values of , , and , we cannot find this value. Therefore, we cannot know the vertex.

III) The -intercept is the point at which ; by substitution, it can be found to be at . is unknown, so the -intercept cannot be found.

IV) The -intercept(s), if any, are the point(s) at which $f(x)= Ax^{2}+ Bx+ C = 0$ . This is solvable using the quadratic formula

$x= \frac{-B \pm \sqrt{B^{2}-4AC}}{2A}$

Since , this can be rewritten and simplified as follows:

$x= \frac{-A \pm \sqrt{A^{2}-4A \cdot A}}{2A}$

$x= \frac{-A \pm \sqrt{A^{2}-4A^{2}}}{2A}$

$x= \frac{-A \pm \sqrt{ -3A^{2}}}{2A}$

$x= \frac{-A \pm A \sqrt{ -3 }}{2A}$

$x= \frac{-1 \pm \sqrt{ -3 }}{2 }$

However, since has no real square root, $f(x)= Ax^{2}+ Bx+ C = 0$ has no real solutions, and its graph has no -intercepts.

V) The line of symmetry has equation $x=-\frac{B}{2A}$ . When exploring the vertex, we found that this value is equal to $- \frac{1}{2}$ , so the line of symmetry is the line of the equation $x=- \frac{1}{2}$ .

The correct response is I, IV, and V only.

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Question

Which of the following equations has as its graph a vertical parabola with line of symmetry ?

Answer

The graph of $f(x)= Ax^{2}+Bx+ C$ has as its line of symmetry the vertical line of the equation

$x=- \frac{B}{2A}$

Since in each choice, we want to find such that

$7=- \frac{B}{2A} = - \frac{B}{2(7)} = - \frac{B}{14}$

$- \frac{B}{14} = 7$

$B = 7\cdot (-14)=-98$

so the correct choice is $f(x)=7x^{2}-98x+140$ .

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Question

Which of the following equations can be graphed with a vertical parabola with exactly one -intercept?

Answer

The graph of $f(x)= Ax^{2}+Bx+ C$ has exactly one -intercept if and only if

$Ax^{2}+Bx+ C = 0$

has exactly one solution - or equivalently, if and only if

$B^{2}-4AC = 0$

Since in all three equations, , we find the value of that makes this statement true by substituting and solving:

$(12)^{2}-4(3)C = 0$

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

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Question

Which of the following equations has as its graph a concave-right horizontal parabola?

Answer

A horizontal parabola has as its equation, in standard form,

$x= Ay^{2}+ By+ C$ ,

with real, nonzero.

Its orientation depends on the sign of . In the equation of a concave-right parabola, is positive, so the correct choice is $x=7y^{2}- 4y+13$ .

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Question

The graphs of the functions and have the same -intercept.

If we define $f(x) = 2x^{2}+ 4x+7$ , which of the following is a possible definition of ?

Answer

The -coordinate of the -intercept of the graph of a function of the form $f(x)= Ax^{2}+Bx+C$ - a quadratic function - is the point . Since $f(x)= A \cdot 0^{2}+B \cdot 0 +C = C$ , the -intercept is at the point .

Because of this, the graph of $f(x) = 2x^{2}+ 4x+7$ has its -intercept at . Among the other choices, only $g(x) = 4x^{2}+ 8x+7$ has a graph with its -intercept also at .

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Question

Give the set of intercepts of the graph of the function $f(x)= \frac{1}{7} (x-7)^{2}$ .

Answer

The -intercepts, if any exist, can be found by setting :

$f(x)= \frac{1}{7} (x-7)^{2} = 0$

$\frac{1}{7} (x-7)^{2} \cdot 7 = 0\cdot 7$

$(x-7)^{2} = 0$

The only -intercept is .

The -intercept can be found by substituting 0 for :

$f(0)= \frac{1}{7} (0-7)^{2} = \frac{1}{7} ( -7)^{2}= \frac{1}{7} \cdot 49 = 7$

The -intercept is .

The correct set of intercepts is $\left { (7,0), (0,7) \right }$ .

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Question

The parabolas of the functions and on the coordinate plane have the same vertex.

If we define $f(x) = \frac{4}{5}(x-9)^{2}+ 6$ , which of the following is a possible equation for ?

Answer

The eqiatopm of is given in the vertex form

$f(x) =a(x-h)^{2}+ k$ ,

so the vertex of its parabola is . The graphs of and are parabolas with the same vertex, so they must have the same values for and .

For the function $f(x) = \frac{4}{5}(x-9)^{2}+ 6$ , and .

Of the five choices, the only equation of that has these same values, and that therefore has a parabola with the same vertex, is $g(x) = \frac{3}{5}(x-9)^{2}+ 6$ .

To verify, graph both functions on the same grid.

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Question

Give the -coordinate of a point of intersection of the graphs of the functions

$f(x) = x^{2}- 4x+ 7$

and

.

Answer

The system of equations can be rewritten as

$y= x^{2}- 4x+ 7$

.

We can set the two expressions in equal to each other and solve:

$x^{2}- 4x+ 7 = 4x$

$x^{2}- 8x+ 7 =0$

We can substitute back into the equation , and see that either or . The latter value is the correct choice.

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Question

The graphs of the functions and have the same pair of -intercepts.

If we define $g(x) = x^{2}- 7x +10$ , which of the following is a possible definition of ?

Answer

The -intercepts of the parabola

$g(x) = x^{2}- 7x +10$

can be determined by setting and solving for :

$x^{2}- 7x +10 = 0$

or

The intercepts of the parabola are and .

We can check each equation to see whether these two ordered pairs satisfy them.

$f(x) = x^{2}- 7x +20$ :

$f(2) = 2^{2}- 7 \cdot 2 +20 = 4-14+20 = 10 e 0$

$f(x) = - x^{2}- 7x +10$

$f(2) = - 2^{2}- 7 \cdot 2 +10 = -4 -14 + 10 = -8 e 0$

$f(x) = x^{2}- 7x-10$

$f(2) = 2^{2}- 7 \cdot 2-10 = 4 - 14 - 10 =-20 e 0$

$f(x) = 2 x^{2}- 14x +10$

$f(x) = 2 \cdot 2^{2}- 14\cdot 2 +10= 8 - 28+10 = -10 e 0$

Each of these equations has a parabola that does not have as an -intercept. However, we look at

$f(x) = 2 x^{2}- 14x +20$

$f(2) = 2\cdot 2^{2}- 14 \cdot 2 +20= 8 - 28 + 20 = 0$

$f(5) = 2\cdot 5^{2}- 14 \cdot 5 +20= 50 - 70+ 20 = 0$

and are also -intercepts of the graph of this function, so this is the correct choice.

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Question

Give the -coordinate of a point at which the graphs of the equations

$y= x^{2}+ 2x- 7$

and

$y = 2x^{2}- 6x+5$

intersect.

Answer

We can set the two quadratic expressions equal to each other and solve for .

$y= x^{2}+ 2x- 7$ and $y = 2x^{2}- 6x+5$ , so

$2x^{2}- 6x+5 = x^{2}+ 2x- 7$

$2x^{2}- 6x+5 -( x^{2}+ 2x- 7)= x^{2}+ 2x- 7 -( x^{2}+ 2x- 7)$

$2x^{2}- 6x+5 - x^{2}- 2x+ 7=0$

$x^{2}- 8x+12=0$

The -coordinates of the points of intersection are 2 and 6. To find the -coordinates, substitute in either equation:

$y= x^{2}+ 2x- 7$

$y= 2^{2}+ 2 \cdot 2 - 7$

One point of intersection is .

$y= x^{2}+ 2x- 7$

$y= 6^{2}+ 2 \cdot 6 - 7$

The other point of intersection is .

1 is not among the choices, but 41 is, so this is the correct response.

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Question

The graphs of the functions and have the same line of symmetry.

If we define $f(x) = 2x^{2}+ 4x+7$ , which of the following is a possible definition of ?

Answer

The graph of a function of the form $f(x)= Ax^{2}+Bx+C$ - a quadratic function - is a vertical parabola with line of symmetry $x= - \frac{B}{2A}$ .

The graph of the function $f(x) = 2x^{2}+ 4x+7$ therefore has line of symmetry

$x= - \frac{4}{2 (2)}$ , or

We examine all four definitions of to find one with this line of symmetry.

$g(x) = 2x^{2}-4x-7$ :

$x= - \frac{-4}{2 (2)}$ , or

$g(x) = x^{2}+ 4x+7$ :

$x= - \frac{ 4}{2 (1)}$ , or

$g(x) = 2x^{2}+ 8x+7$

$x= - \frac{ 8}{2 (2)}$ , or

$g(x) = 4x^{2}+8x+7$

$x= - \frac{ 8}{2 (4)}$ , or

Since the graph of the function $g(x) = 4x^{2}+8x+7$ has the same line of symmetry as that of the function $f(x) = 2x^{2}+ 4x+7$ , that is the correct choice.

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