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?

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.

Answer

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

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

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Question

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

Give its equation.

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.

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.

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.

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.

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.

Answer

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

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

or

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

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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.

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.

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.

Answer

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 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

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

$\frac{(x-3)^{2}}{16} + \frac{(y-5)^{2}}{9} = 1$

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Question

Give the equation of the above ellipse.

Answer

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 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

$\frac{(x-6)^{2}}{4^{2}} + \frac{(y-2)^{2}}{8^{2}} = 1$

$\frac{(x-6)^{2}}{16} + \frac{(y-2)^{2}}{64} = 1$

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Question

Give the equation of the above ellipse.

Answer

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

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

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

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Question

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

Give its equation.

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.

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.

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.

Answer

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

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

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

$\frac{(x-7)^{2}}{3^{2}} + \frac{(y-6)^{2}}{4^{2}} = 1$ , or

$\frac{(x-7)^{2}}{9} + \frac{(y-6)^{2}}{16} = 1$

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Question

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

Give its equation.

Answer

The standard form of the equation of a horizontal parabola is

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

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 $x=2y^{2}+5y-8$ .

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Question

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

Give its equation.

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?

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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