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

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

Explanation

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.

A vertical parabola on the coordinate plane includes points and .

Give its equation.

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

$y = 2x^{2}+4x-1$

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

$y = x^{2}+4x -8$

$y = x^{2}-4x+8$

Explanation

The standard form of the equation of a vertical parabola is

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

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

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

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

Give its equation.

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

Insufficient information is given to determine the equation.

$y =- 3x^{2}-36x-111$

$y = 3x^{2}+36x+105$

$y =- \frac{3}{8}x^{2}- \frac{9}{4}x - \frac{75}{8}$

Explanation

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:

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

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

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

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

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

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

Give its equation.

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

Insufficient information is given to determine the equation.

$y = \frac{3}{4} x^{2} + 6x+22$

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

$y = \frac{1}{2} x^{2}+4x+18$

Explanation

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:

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

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

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

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

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

Give its equation.

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

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

$y = \frac{3}{2}x^{2}-12x+18$

$y = - \frac{3}{2}x^{2}+12x-18$

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

Explanation

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:

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

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

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

Give its equation.

Insufficient information is given to determine the equation.

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

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

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

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

Explanation

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.

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

Give its equation.

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

Insufficient information is given to determine the equation.

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

$y = \frac{1}{6} x^{2}-\frac{5}{3}x+4$

$y =- \frac{1}{6} x^{2}+\frac{1}{3}x+4$

Explanation

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:

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

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

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

Ellipse 1

Give the equation of the above ellipse.

$\frac{(x-1)^{2}}{25} + \frac{(y-1)^{2}}{9} = 1$

$\frac{(x+1)^{2}}{25} + \frac{(y+1)^{2}}{9} = 1$

$\frac{(x-1)^{2}}{100} + \frac{(y-1)^{2}}{36} = 1$

$\frac{(x+1)^{2}}{100} + \frac{(y+1)^{2}}{36} = 1$

$\frac{(x-1)^{2}}{5} + \frac{(y-1)^{2}}{3} = 1$

Explanation

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

$\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$

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

$\frac{(x-1)^{2}}{5^{2}} + \frac{(y-1)^{2}}{3^{2}} = 1$

$\frac{(x-1)^{2}}{25} + \frac{(y-1)^{2}}{9} = 1$

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

$\frac{(x+4)^{2}}{225} + \frac{(y-9)^{2}}{16} = 1$

Insufficient information is given to determine the equation.

$\frac{(x-4)^{2}}{225} + \frac{(y+9)^{2}}{16} = 1$

$\frac{(x-4)^{2}}{900} + \frac{(y+9)^{2}}{64} = 1$

$\frac{(x+4)^{2}}{900} + \frac{(y-9)^{2}}{64} = 1$

Explanation

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

$\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$

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

$\frac{(x-(-4))^{2}}{15^{2}} + \frac{(y-9)^{2}}{4^{2}} = 1$

$\frac{(x+4)^{2}}{225} + \frac{(y-9)^{2}}{16} = 1$ .

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.