Parabolas - College Algebra

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Question

Which of the following would give a graph of an upward facing parabola?

Answer

Which of the following would give a graph of an upward facing parabola?

Parabolas are created from polynomials where the highest exponent is 2.

To be an upward facing parabola, the squared term must be positive.

The only option that matches these criteria is:

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Question

Find the coordinates of the vertex of this quadratic equation:

Answer

To find the vertex of this parabola use the following formula to find the x-coordinate of its vertex; find the y-coordinate by substituting the x-coordinate into the equation.

$\frac{-b}{2a}$

$\frac{-24}{2(6)}=\boldsymbol{-2}$

To find the y-coordinate, substitute -2 back into the quadratic equation:

$6x^2+24x+18\rightarrow 6(-2)^2+24(-2)+18=\boldsymbol{-6}$

The vertex is $\boldsymbol{(-2,-6)}$ .

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Question

What is the multiplicity of the root of a quadratic equation with a discriminant equal to 0?

Answer

Quadratic equations whose discriminants are equal to zero have one repeated root (solution). Because this root appears twice in the quadratic equation, it has a multiplicity of 2. The number of times a factor appears in a polynomial, such as quadratic, is its multiplicity.

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Question

How many -intercepts does the graph of the function

have?

Answer

The graph of a quadratic function $f(x) = ax^{2}+ bx + c$ has an -intercept at any point at which , so, first, set the quadratic expression equal to 0:

The number of -intercepts of the graph is equal to the number of real zeroes of the above equation, which can be determined by evaluating the discriminant of the equation, $b^{2} -4ac$ . Set , and evaluate:

$b^{2} -4ac = 12^{2} - 4 (3)(8) = 14 4 - 96 = 48$

The discriminant is positive, so the equation has two solutions, both of which are real. Consequently, the graph of the function has two -intercepts

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Question

A baseball is thrown off the roof of a building 320 feet high at an initial upward speed of 80 feet per second; the height of the baseball relative to the ground is modeled by the function

$h(t) = -16t^{2}+ 80t + 320$

How long does it take for the baseball to reach its highest point (nearest tenth of a second)?

Answer

The highest point of the ball is the vertex of the ball's parabolic path, so to find the number of seconds that is takes to reach this point, it is necessary to find the first coordinate of the vertex of the parabola of the graph of the function

$h(t) = -16t^{2}+ 80t + 320$

The parabola of the graph of

$h(t) =at^{2}+ bt+ c$

has as its ordinate, or -coordinate,

$t = -\frac{b}{2a}$ ,

so, setting ,

$t = -\frac{80}{2 (-16)} = -\frac{80}{-32} = 2.5$ ,

This is the time in seconds that it takes the ball to reach the highest point of its path.

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Question

Define a function $f(x) = -4x^{2}+ 8x -2$ .

Which of the following is the -coordinate of the -intercept of its graph?

Answer

The -intercept of the graph of a function is the point at which it crosses the -axis; its -coordinate is 0, so its -coordinate is .

$f(x) = -4x^{2}+ 8x -2$ ,

so, by setting ,

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

The -intercept is .

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Question

Try without a calculator.

The graph of the function

is a parabola. Which choice correctly gives its concavity?

Answer

The direction of the concavity of the parabola of the function

$f(x) = ax^{2}+ bx + c$

is either upward or downward depending entirely on the sign of , the coefficient of $x^{2}$ . This coefficient, , is negative; the parabola is concave downward.

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Question

Give the vertex of the parabola of the equation

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

Answer

The parabola of the equation $x = ay^{2}+ by+ c$ has its vertex at a point with -coordinate $- \frac{b}{2a}$ ; set and in this formula to get

$- \frac{8}{2(1)} = -4$

Substitute this for in the equation to obtain the corresponding -coordinate:

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

$x= ( -4)^{2}+ 8( -4) - 12 = 16 + (-32 )- 12 = -28$

The vertex is at $\left ( -28 , -4\right )$ .

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Question

Give the equation of the directrix of the parabola of the equation

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

Answer

The parabola in question is a horizontal parabola. Its equation is in the standard form

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

Before the directrix can be found, it is necessary to find the vertex . This is located at the point with ordinate

$y= -\frac{b}{2a} = - \frac{4}{2(2)} = -1$

and abscissa

$x=2 (-1)^{2}+ 4(-1) - 8$

That is, the vertex is at .

The directrix of the parabola is the line of the equation $x = h - \frac{1}{4a}$ , which is

$x = -10- \frac{1}{4(2)}$

$x = -10- \frac{1}{8}$

$x = - \frac{81}{8}$

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Question

Give the coordinates of the focus of the parabola of the equation

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

Answer

The parabola in question is a vertical parabola. Its equation is in the standard form

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

Before the focus can be found, it is necessary to find the vertex . This is located at the point with abscissa

$x = -\frac{b}{2a} = -\frac{4}{2(4)}= -\frac{1}{2}$ .

Substitute this for to find the ordinate:

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

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

$= 4 \left ( \frac{1}{4} \right ) + 4 \left ( -\frac{1}{2} \right )+ 8$

The vertex of the parabola is $\left ( -\frac{1}{2} , 7 \right )$ .

The focus of a vertical parabola is located at the point

$\left ( h, k + \frac{1}{4a} \right )$ .

Setting $h = -\frac{1}{2} , k = 7, a = 4$ , the point has coordinates

$\left ( -\frac{1}{2}, 7 + \frac{1}{4 (4)} \right )$ .

$7 + \frac{1}{4 (4)} = 7 + \frac{1}{16} = \frac{113}{6}$ , so the focus is at

$\left ( -\frac{1}{2}, \frac{113}{16} \right )$ .

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