How to find the equation of a curve - SSAT Upper Level Quantitative

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Question

If the -intercept of the line is and the slope is , which of the following equations best satisfies this condition?

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Answer

Write the slope-intercept form.

The point given the x-intercept of 6 is .

Substitute the point and the slope into the equation and solve for the y-intercept.

Substitute the y-intercept back to the slope-intercept form to get your equation.

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Question

A vertical parabola on the coordinate plane includes points and .

Give its equation.

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Answer

The standard form of the equation of a vertical parabola is

If the values of and from each ordered pair are substituted in succession, three equations in three variables are formed:

$a\cdot $(-2)^{2}$+b(-2)+c = 23$

$a\cdot $1^{2}$+b \cdot 1+c = 5$

$a\cdot $3^{2}$+b \cdot 3+c =13$

The system

can be solved through the elimination method.

First, multiply the second equation by and add to the third:

$\underline{ -a-b-c ; ; ; =- 5 }$

Next, multiply the second equation by and add to the first:

$\underline{ -a-b-c ; =- 5 }$

Which can be divided by 3 on both sides to yield

Now solve the two-by-two system

by substitution:

Back-solve:

Back-solve again:

The equation of the parabola is therefore

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

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Question

A vertical parabola on the coordinate plane has vertex and -intercept .

Give its equation.

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Answer

The equation of a vertical parabola, in vertex form, is

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

where is the vertex. Set :

$y = $a[x-(-4)]^{2}$+10$

$y = $a(x+4)^{2}$+10$

To find , use the -intercept, setting :

$-2 = $a(0+4)^{2}$+10$

$-2 = a\cdot $4^{2}$+10$

$a =- $\frac{3}{4}$$

The equation, in vertex form, is $y =- $$\frac{3}{4}$(x+4)^{2}$+10$ ; in standard form:

$y = $-$\frac{3}{4}$(x+4)^{2}$+10$

$y =- $$\frac{3}{4}$(x^{2}$ + 8x+16)+10$

$y = -$\frac{3}{4}$ $x^{2}$- 6x-12+10$

$y = -$\frac{3}{4}$ $x^{2}$- 6x-2$

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Question

A vertical parabola on the coordinate plane has vertex ; one of its -intercepts is .

Give its equation.

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Answer

The equation of a vertical parabola, in vertex form, is

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

where is the vertex. Set :

$y = $a[x-(-3)]^{2}$+ (-6)$

$y = $a(x+3)^{2}$-6$

To find , use the known -intercept, setting :

$0 = $a(-7+3)^{2}$-6$

$0 = $a(-4)^{2}$-6$

$a = $\frac{6}{16}$ = $\frac{3}{8}$$

The equation, in vertex form, is $y = $$\frac{3}{8}$(x+3)^{2}$-6$ ; in standard form:

$y = $$\frac{3}{8}$(x+3)^{2}$-6$

$y = $$\frac{3}{8}$(x^{2}$+6x+9)-6$

$y = $$\frac{3}{8}$x^{2}$+ $\frac{9}{4}$x+ $\frac{27}{8}$-6$

$y = $$\frac{3}{8}$x^{2}$+ $\frac{9}{4}$x+ $\frac{27}{8}$- $\frac{48}{8}$$

$y = $$\frac{3}{8}$x^{2}$+ $\frac{9}{4}$x - $\frac{21}{8}$$

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Question

A vertical parabola on the coordinate plane has -intercept ; its only -intercept is .

Give its equation.

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Answer

If a vertical parabola has only one -intercept, which here is , that point doubles as its vertex as well.

The equation of a vertical parabola, in vertex form, is

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

where is the vertex. Set :

$y = $a(x-6)^{2}$+0$

$y = $a(x-6)^{2}$$

To find , use the -intercept, setting :

$4= $a(0-6)^{2}$$

$4= a( $-6)^{2}$$

$a = $\frac{4}{36}$ = $\frac{1}{9}$$

The equation, in vertex form, is $y = $$\frac{1}{9}$(x-6)^{2}$$ . In standard form:

$y = $$\frac{1}{9}$(x-6)^{2}$$

$y = $$\frac{1}{9}$(x^{2}$-12+36)$

$y = $\frac{1}{9}$ $x^{2}$-$\frac{4}{3}$x+4$

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Question

A vertical parabola on the coordinate plane has -intercept ; one of its -intercepts is .

Give its equation.

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Answer

The equation of a vertical parabola, in standard form, is

$y = $ax^{2}$+ bx + c$

for some real .

is the -coordinate of the -intercept, so , and the equation is

$y = $ax^{2}$+ bx + 6$

Set :

$0 = a\cdot $2^{2}$+ b \cdot 2 + 6$

However, no other information is given, so the values of and cannot be determined for certain. The correct response is that insufficient information is given.

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Question

A vertical parabola on the coordinate plane has -intercepts and , and passes through .

Give its equation.

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Answer

A vertical parabola which passes through and has as its equation

To find , substitute the coordinates of the third point, setting :

$-2 = a \cdot 1 \cdot (-3)$

$a = $\frac{-2}{-3}$= $\frac{2}{3}$$

The equation is $y = $\frac{2}{3}$(x-2)(x-6)$ ; expand to put it in standard form:

$y = $$\frac{2}{3}$(x^{2}$-8x+12)$

$y = $$\frac{2}{3}$x^{2}$-$\frac{16}{3}$x+8$

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Question

An ellipse on the coordinate plane has as its center the point . It passes through the points and . Give its equation.

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Answer

The equation of the ellipse with center , horizontal axis of length , and vertical axis of length is

The center is , so and .

To find , note that one endpoint of the horizontal axis is given by the point with the same -coordinate through which it passes, namely, . Half the length of this axis, which is , is the difference of the -coordinates, so . Similarly, to find , note that one endpoint of the vertical axis is given by the point with the same -coordinate through which it passes, namely, . Half the length of this axis, which is , is the difference of the -coordinates, so .

The equation is

or

.

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Question

A vertical parabola on the coordinate plane shares one -intercept with the line of the equation , and the other with the line of the equation . It also passes through . Give the equation of the parabola.

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Answer

First, find the -intercepts—the points of intersection with the -axis—of the lines by substituting 0 for in both equations.

is the -intercept of this line.

is the -intercept of this line.

The parabola has -intercepts at and , so its equation can be expressed as

for some real . To find it, substitute using the coordinates of the third point, setting :

$-4 = a \cdot 3 \cdot 2$

$a = $\frac{-4}{6}$ = -$\frac{2}{3}$$ .

The equation is $y = -$\frac{2}{3}$(x-3)(x-4)$ , which, in standard form, can be rewritten as:

$y = $-$\frac{2}{3}$(x^{2}$-7x+12)$

$y = $-$\frac{2}{3}$x^{2}$+\frac{14}{3}$x-8$

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Question

The -intercept and the only -intercept of a vertical parabola on the coordinate plane coincide with the -intercept and the -intercept of the line of the equation . Give the equation of the parabola.

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Answer

To find the -intercept, that is, the point of intersection with the -axis, of the line of equation , set and solve for :

The -intercept is .

The -intercept can be found by doing the opposite:

The -intercept is .

The parabola has these intercepts as well. Also, since the vertical parabola has only one -intercept, that point doubles as its vertex as well.

The equation of a vertical parabola, in vertex form, is

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

where is the vertex. Set :

$y = $a(x-10)^{2}$+0$

$y = $a(x-10)^{2}$$

for some real . To find it, use the -intercept, setting

$6 = $a(0-10)^{2}$$

$6 = a( $-10)^{2}$$

$a = $\frac{6}{100}$ = $\frac{3}{50}$$

The parabola has equation $y =\frac{3}{50}$ $(x-10)^{2}$$ , which is rewritten as

$y =\frac{3}{50}$ $(x^{2}$-20x+100)$

$y = $\frac{3}{50}$ $x^{2}$-$\frac{6}{5}$x+6$

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Question

Give the equation of the above ellipse.

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Answer

The equation of the ellipse with center , horizontal axis of length , and vertical axis of length is

The ellipse has center , horizontal axis of length 8, and vertical axis of length 6. Therefore,

, $a = $\frac{8}{2}$ = 4$ , and $b = $\frac{6}{2}$ = 3$ .

The equation of the ellipse is

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Question

Give the equation of the above ellipse.

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Answer

The equation of the ellipse with center , horizontal axis of length , and vertical axis of length is

The ellipse has center , horizontal axis of length 8, and vertical axis of length 16. Therefore,

, $a = $\frac{8}{2}$ = 4$ , and $b = $\frac{16}{2}$ = 8$ .

The equation of the ellipse is

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Question

Give the equation of the above ellipse.

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Answer

The equation of the ellipse with center , horizontal axis of length , and vertical axis of length is

The ellipse has center , horizontal axis of length 10, and vertical axis of length 6. Therefore,

, $a = $\frac{10}{2}$ = 5$ , and $b = $\frac{6}{2}$ = 3$ .

The equation of the ellipse is

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Question

A horizontal parabola on the coordinate plane as its only -intercept; its -intercept is .

Give its equation.

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Answer

If a horizontal parabola has only one -intercept, which here is , that point doubles as its vertex as well.

The equation of a horizontal parabola, in vertex form, is

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

where is the vertex. Set :

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

$x = $a(y-4)^{2}$$

To find , use the -intercept, setting :

$6 = $a(0-4)^{2}$$

$6 = a( $-4)^{2}$$

$a = $\frac{6}{16}$ = $\frac{3}{8}$$

The equation, in vertex form, is $x = $$\frac{3}{8}$(y-4)^{2}$$ . In standard form:

$x = $$\frac{3}{8}$(y^{2}$-8y+16)$

$x = $\frac{3}{8}$ $y^{2}$-3y+6$

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Question

A horizontal parabola on the coordinate plane has -intercept ; one of its -intercepts is .

Give its equation.

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Answer

The equation of a horizontal parabola, in standard form, is

$x = $ay^{2}$+ by + c$

for some real .

is the -coordinate of the -intercept, so , and the equation is

$x = $ay^{2}$+ by + 6$

Set :

$0 = a\cdot $2^{2}$+ b \cdot 2 + 6$

However, no other information is given, so the values of and cannot be determined for certain. The correct response is that insufficient information is given.

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Question

A horizontal parabola on the coordinate plane has vertex ; one of its -intercepts is .

Give its equation.

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Answer

The equation of a horizontal parabola, in vertex form, is

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

where is the vertex. Set :

$x = $a[y-(-6)]^{2}$+ (-3)$

$x = $a(y+6)^{2}$-3$

To find , use the known -intercept, setting :

$0 = $a(4+6)^{2}$-3$

$0 = a\cdot $10^{2}$-3$

$a = $\frac{3}{100}$$

The equation, in vertex form, is $x = $$\frac{3}{100}$(y+6)^{2}$-3$ ; in standard form:

$x = $$\frac{3}{100}$(y^{2}$+12+36)-3$

$x = $\frac{3}{100}$ $y^{2}$+\frac{9}{25}$ y+\frac{27}{25}$ - $\frac{75}{25}$$

$x = $\frac{3}{100}$ $y^{2}$+\frac{9}{25}$ y-$\frac{48}{25}$$

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Question

An ellipse passes through points .

Give its equation.

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Answer

The equation of the ellipse with center , horizontal axis of length , and vertical axis of length is

and are the endpoints of a horizontal line segment with midpoint

$\left ( $\frac{4+10}{2}$, $\frac{6+6}{2}$ \right )$ , or

and length .

and are the endpoints of a vertical line segment with midpoint

$\left ( $\frac{7+7}{2}$, $\frac{2+10}{2}$ \right )$ , or

and length

Because their midpoints coincide, these are the endpoints of the horizontal axis and vertical axis, respectively, of the ellipse, and the common midpoint is the center.

Therefore,

and ;

and ; consequently and .

The equation of the ellipse is

, or

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Question

A horizontal parabola on the coordinate plane includes points , and .

Give its equation.

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Answer

The standard form of the equation of a horizontal parabola is

If the values of and from each ordered pair are substituted in succession, three equations in three variables are formed:

$a \cdot $(-1)^{2}$+b \cdot (-1)+ c = -11$

$a \cdot $(1)^{2}$+b \cdot 1+ c = -1$

$a \cdot $(2)^{2}$+b \cdot 2+ c = 10$

The three-by-three linear system

can be solved by way of the elimination method.

can be found first, by multiplying the first equation by and add it to the second:

$\underline{-a+b - c = 11}$

$2b \div 2 = 10 \div 2$

Substitute 5 for in the last two equations to form a two-by-two linear system:

$4a+2 \cdot 5+ c = 10$

The system

can be solved by way of the substitution method;

$-3a\div (-3) = -6 \div (-3)$

Substitute 2 for in the top equation:

The equation is .

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Question

A vertical parabola on the coordinate plane has -intercepts and , and passes through .

Give its equation.

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Answer

A horizontal parabola which passes through and has as its equation

.

To find , substitute the coordinates of the third point, setting :

$a = $\frac{3}{32}$$

The equation is therefore $x = $\frac{3}{32}$(y-2)(y-6)$ , which is, in standard form:

$x = $$\frac{3}{32}$(y^{2}$-8y+12)$

$x = $\frac{3}{32}$ $y^{2}$- $\frac{3}{4}$ y+ $\frac{9}{8}$$

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Question

If the -intercept of the line is and the slope is , which of the following equations best satisfies this condition?

Tap to reveal answer

Answer

Write the slope-intercept form.

The point given the x-intercept of 6 is .

Substitute the point and the slope into the equation and solve for the y-intercept.

Substitute the y-intercept back to the slope-intercept form to get your equation.

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