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Precalculus : Pre-Calculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Solve A System Of Quadratic Equations

Find the coordinate of intersection, if possible: y=x^2+4 and y=-x^2+4 .

Possible Answers:

(4,0)

$(0,\pm4)$

(-4,0)

(0,4)

$\textup{There is no intersection.}$

Correct answer:

(0,4)

Explanation:

To solve for x and y, set both equations equal to each other and solve for x.

-x^2+4=x^2+4

2x^2 =0

x=0

Substitute x=0 into either parabola.

y=0^2+4 = 4

The coordinate of intersection is (0,4) .

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Example Question #3 : Solve A System Of Quadratic Equations

Find the intersection(s) of the two parabolas: y=2x^2+3x+4 , y=-4x^2+4

Possible Answers:

$\left(\frac{1}{2},3\right)$

$(0,4) \textup{ and }\left(\frac{1}{2},3\right)$

$(0,4) \textup{ and }\left(-\frac{1}{2},3\right)$

$\left(-\frac{1}{2},3\right)$

(0,4)

Correct answer:

$(0,4) \textup{ and }\left(-\frac{1}{2},3\right)$

Explanation:

Set both parabolas equal to each other and solve for x.

2x^2+3x+4=-4x^2+4

6x^2+3x=0

3x(2x+1)=0

$x=0,-\frac{1}{2}$

Substitute both values of into either parabola and determine .

y=-4(0)^2+4= 4

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

The coordinates of intersection are:

(0,4) and $\left(-\frac{1}{2},3\right)$

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Example Question #1 : Solve A System Of Quadratic Equations

Find the points of intersection:

y = x^2 + 40x - 6 ; y = -x^2 + 19 x + 5

Possible Answers:

(-11, -567) ; (-0.5, -25.75)

(11, 555) ; (0.5 , 14.25)

(-11, -567); (0.5, 9.05)

(-0.5, -25.75) ; (-11, -325)

(-11, -325); (0.5, 14.25)

Correct answer:

(-11, -325); (0.5, 14.25)

Explanation:

To solve, set both equations equal to each other:

x^2 + 40x - 6 = -x^2 + 19x + 5

To solve as a quadratic, combine like terms by adding/subtracting all three terms from the right side to the left side:

x^2 + x^2 +40x - 19x - 6 - 5 = 0

This simplifies to

2x^2 + 21 x - 11 = 0

Solving by factoring or the quadratic formula gives the solutions and 0.5 .

Plugging each into either original equation gives us:

(-11)^2 + 40(-11 ) - 6 = 121 - 440 - 6 = -325

(0.5) ^ 2 + 40 (0.5) - 6 = 14.25

Our coordinate pairs are (-11, -325) and (0.5, 14.25 ) .

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Example Question #2 : Solve A System Of Quadratic Equations

Give the coordinate pairs that satisfy the system of equations.

$\begin{matrix} y = 5x^2 - 10x + 6 \\ y = 2x^2 + 9 x + 20 \end{matrix}$

Possible Answers:

$(7, 181); \left(\frac{ 2}{3} , 1. \overline{5}\right)$

$\left( -\frac{ 2}{3 } , 14. \overline{8}\right); (7 , 181)$

$\left(- \frac{ 2}{3} , 14.\overline{8}\right); (-7, 41)$

$\left(- \frac{ 2}{3} , 14. \overline{8 }\right) ; (-7, 321)$

$(-7, 321) ; \left(\frac{2}{3} , 12. \overline{8}\right)$

Correct answer:

$\left( -\frac{ 2}{3 } , 14. \overline{8}\right); (7 , 181)$

Explanation:

To solve, set the two quadratics equal to each other and then combine like terms:

5 x ^ 2 - 10 x + 6 = 2x ^2 + 9x + 20 subtract everything on the right from both sides to combine like terms.

5x^2 - 2x^2 - 10x - 9 x +6 - 20 = 0

3x^2 - 19x - 14=0

Solving by factoring or using the quadratic formula gives us the solutions $-\frac{2}{3}$ and .

To find the y-coordinates, plug these into either equation:

$5\left( -\frac{ 2}{3} \right)^2 -10 \left(-\frac{ 2}{3}\right) + 6 = 14.\overline{8}$

2(7)^2 + 9(7) + 20 = 181

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Example Question #2 : Solve A System Of Quadratic Equations

Give the , coordinate pairs that satisfy the two equations.

$\begin{matrix} y = x^2 - 7x - 10 \\ y +2x^2 + 18x = 10 \end{matrix}$

Possible Answers:

$(-5, 50 ) ; \left(\frac{ 4}{3} , -17.\overline{5}\right)$

$\left(\frac{ 3}{4} , -14.7\right); (-5, 50)$

$(5, -20) ; \left(\frac{ 4}{3} , -17.\overline{5}\right)$

$\left(- \frac{ 4}{3}, 1.\overline{1}\right) ; (5, -20)$

$(5, -130) ; \left(\frac{3}{4} , -14.7\right)$

Correct answer:

$(-5, 50 ) ; \left(\frac{ 4}{3} , -17.\overline{5}\right)$

Explanation:

To solve, first re-write the second one so that y is isolated on the left side:

y = -2x ^ 2 - 18 x + 10

Now set the two quadratics equal to each other:

x^ 2 - 7 x - 10 = -2x^2 - 18x + 10 add/subtract all of the terms from the right side so that this is a quadratic equal to zero.

x^2 + 2x^2 - 7x + 18 x -10 -10 = 0 combine like terms.

3x^2 + 11x -20 = 0

Using the quadratic formula or by factoring, we get the two solutions and $\frac{ 4}{3}$ .

To get the y-coordinates, plug these numbers into either function:

y = (-5)^2 - 7 (-5) - 10 = 25+35 - 10 = 50

$y = \left(\frac{4}{3}\right)^2 - 7\left(\frac{ 4}{3} \right) - 10 = \frac {16}{9} - \frac{84}{9} - \frac{90}{9} = \frac{-158}{9 } = -17.\overline{5}$

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Example Question #4 : Solve A System Of Quadratic Equations

Find the coordinate pairs satisfying both polynomials:

$\begin{matrix} y = 4x^3 + 3x^2 -10x - 5 \\ y = 4x^3 + 2x^2 -8x + 10 \end{matrix}$

Possible Answers:

(3, 112) ; (5, 520 )

(-5, -400) ; (-3, -56)

(3, 112) ; (-5, -380)

(3, 100) ; (-5, -400)

(-3, -56); (5, 520)

Correct answer:

(-3, -56); (5, 520)

Explanation:

To solve, set the two polynomials equal to each other:

4x^3 + 3x^2 -10x - 5 = 4x^3 + 2x^2 -8x + 10 add/subtract all of the terms from the right side from both sides.

4x^3 - 4x^3 + 3x^2 - 2x^2 -10x + 8 x - 5 -10 = 0 combine like terms.

x^2 -2x -15 = 0

Solving with the quadratic formula or by factoring gives us the solutions 5 and -3.

To get the y-coordinates, plug these numbers into either of the original equations:

4(5)^3 +3(5)^2 -10(5) - 5 = 500 + 75 - 50 - 5 = 520

4(-3)^3 + 2(-3) ^ 2 - 8 (-3 ) + 10 = -108 + 18 + 24 + 10 = -56

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Example Question #1 : Polar Equations Of Conic Sections

Given the polar equation, determine the conic section:

$r=\frac{2}{1-2\sin(\theta)}$

Possible Answers:

Hyperbola

Parabola

Ellipse

Correct answer:

Hyperbola

Explanation:

Recall that the polar equations of conic sections can come in the following forms:

$r=\frac{ed}{1\pm e \cos(\theta)}$

$r=\frac{ed}{1\pm e \sin(\theta)}$ , where is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

0<e<1 will give an ellipse.

e=1 will give a parabola.

e>1 will give a hyperbola.

Now, for the given conic section, e>1 so it must be a hyperbola.

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Example Question #2 : Polar Equations Of Conic Sections

Given the polar equation, determine the conic section:

$r=\frac{6}{1-\frac{1}{2}\sin(\theta)}$

Possible Answers:

Parabola

Ellipse

Hyperbola

Correct answer:

Ellipse

Explanation:

Recall that the polar equations of conic sections can come in the following forms:

$r=\frac{ed}{1\pm e \cos(\theta)}$

$r=\frac{ed}{1\pm e \sin(\theta)}$ , where is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

0<e<1 will give an ellipse.

e=1 will give a parabola.

e>1 will give a hyperbola.

Now, for the given conic section, 0<e<1 so it must be an ellipse.

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Example Question #1 : Identify The Conic With A Given Polar Equation

Given the polar equation, determine the conic sectioN:

$r=\frac{8}{4-\sin\theta}$

Possible Answers:

Ellipse

Parabola

Hyperbola

Correct answer:

Ellipse

Explanation:

Recall that the polar equations of conic sections can come in the following forms:

$r=\frac{ed}{1\pm e \cos(\theta)}$

$r=\frac{ed}{1\pm e \sin(\theta)}$ , where is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

0<e<1 will give an ellipse.

e=1 will give a parabola.

e>1 will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by .

$r=\frac{8}{4-\sin\theta}=\frac{2}{1-\frac{1}{4}\sin\theta}$

Now, for the given conic section, 0<e<1 so it must be an ellipse.

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Example Question #4 : Polar Equations Of Conic Sections

Given the polar equation, determine the conic section:

$r=\frac{6}{6-6\cos\theta}$

Possible Answers:

Parabola

Hyperbola

Ellipse

Correct answer:

Parabola

Explanation:

Recall that the polar equations of conic sections can come in the following forms:

$r=\frac{ed}{1\pm e \cos(\theta)}$

$r=\frac{ed}{1\pm e \sin(\theta)}$ , where is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

0<e<1 will give an ellipse.

e=1 will give a parabola.

e>1 will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by .

$r=\frac{6}{6-6\cos\theta}=\frac{1}{1-\cos\theta}$

Now, for the given conic section, e=1 so it must be a parabola.

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #2 : Solve A System Of Quadratic Equations

Example Question #3 : Solve A System Of Quadratic Equations

Example Question #1 : Solve A System Of Quadratic Equations

Example Question #2 : Solve A System Of Quadratic Equations

Example Question #2 : Solve A System Of Quadratic Equations

Example Question #4 : Solve A System Of Quadratic Equations

Example Question #1 : Polar Equations Of Conic Sections

Example Question #2 : Polar Equations Of Conic Sections

Example Question #1 : Identify The Conic With A Given Polar Equation

Example Question #4 : Polar Equations Of Conic Sections

All Precalculus Resources

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