Parabolic Functions - Algebra II

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Question

Write a quadratic equation having $\left ( -3,2 \right )$ as the vertex (vertex form of a quadratic equation).

Answer

The vertex form of a quadratic equation is given by

$\left ( x - h \right )^{2} + k$

Where the vertex is located at $\left ( h,k \right )$

giving us $\left ( x + 3 \right )^{2} +2$ .

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Question

What is the minimum possible value of the expression below?

$3x^{2}+6x-10$

Answer

We can determine the lowest possible value of the expression by finding the -coordinate of the vertex of the parabola graphed from the equation $y=3x^{2}+6x-10$ . This is done by rewriting the equation in vertex form.

$3x^{2}+6x-10$

$3(x^{2}+2x)-10$

$3(x^{2}+2x+1)-3* 1-10$

$3(x+1)^{2}-13$

The vertex of the parabola $y=3(x+1)^{2}-13$ is the point .

The parabola is concave upward (its quadratic coefficient is positive), so represents the minimum value of . This is our answer.

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Question

Which of the following functions represents a parabola?

Answer

A parabola is a curve that can be represented by a quadratic equation. The only quadratic here is represented by the function $f(x) = x^{2} - 9$ , while the others represent straight lines, circles, and other curves.

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Question

What is the equation of a parabola with vertex and -intercept ?

Answer

From the vertex, we know that the equation of the parabola will take the form $y = a (x - 4) ^{2}+ 1$ for some .

To calculate that , we plug in the values from the other point we are given, , and solve for :

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

$-7= a \cdot (0 - 4)^{2} + 1$

$-7= a \cdot (- 4)^{2} + 1$

$a = -\frac{1}{2}$

Now the equation is $y = -\frac{1}{2} (x - 4)^{2} + 1$ . This is not an answer choice, so we need to rewrite it in some way.

Expand the squared term:

$y = -\frac{1}{2} (x^{2} -8x + 16) + 1$

Distribute the fraction through the parentheses:

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

Combine like terms:

$y = -\frac{1}{2} x^{2} +4x - 7$

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Question

Give the minimum value of the function $f(x) = 2 x^{2} - 5x + 9$ .

Answer

This is a quadratic function. The -coordinate of the vertex of the parabola can be determined using the formula $x = - \frac{b}{2a}$ , setting :

$x = - \frac{b}{2a} = - \frac{-5}{2 \cdot 2} =\frac{5}{4}$

Now evaluate the function at $x = \frac{5}{4}$ :

$f(x) = 2 x^{2} - 5x + 9$

$f \left ( \frac{5}{4} \right ) = 2 \left ( \frac{5}{4} \right )^{2} - 5 \left ( \frac{5}{4} \right )+ 9$

$= 2 \left (\frac{25}{16} \right ) - 5 \left ( \frac{5}{4} \right )+ 9$

$= \frac{25}{8} - \frac{25}{4} + 9$

$= \frac{25}{8} - \frac{50}{8} + \frac{72}{8} =\frac{47}{8} =5\frac{7}{8}$

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Question

What are the -intercepts of the equation?

$y=\frac{x^2-9}{x^2-16}$

Answer

To find the x-intercepts of the equation, we set the numerator equal to zero.

$\sqrt{9}=\sqrt{x^2}$

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Question

Find the coordinates of the vertex of this quadratic function:

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

Answer

Vertex of quadratic equation $y=ax^{2}+bx+c$ is given by .

$h=-\frac{b}{2a}, k=c-\frac{b^{2}}{4a}$

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

$h=-\frac{6}{2\times (-2)}=\frac{3}{2},$

$k=-5-\frac{6^{2}}{4\times (-2)}=-5+\frac{9}{2}=-\frac{1}{2}$ ,

so the coordinate of vertex is $\left(\frac{3}{2}, -\frac{1}{2}\right)$ .

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Question

What are the x-intercepts of the graph of $y=8x^{2}+2x-3$ ?

Answer

Assume y=0,

$y=8x^{2}+2x-3=0$

$x=\frac{1}{2}$ , $x=-\frac{3}{4}$

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Question

Find the vertex of the parabola given by the following equation:

Answer

In order to find the vertex of a parabola, our first step is to find the x-coordinate of its center. If the equation of a parabola has the following form:

Then the x-coordinate of its center is given by the following formula:

$x_{center}=-\frac{b}{2a}$

For the parabola described in the problem, a=-2 and b=-12, so our center is at:

$x_{center}=-\frac{-12}{2(-2)}=-3$

Now that we know the x-coordinate of the parabola's center, we can simply plug this value into the function to find the y-coordinate of the vertex:

So the vertex of the parabola given in the problem is at the point

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Question

Determine the maximum or minimum of .

Answer

To find the max or min of , use the vertex formula and substitute the appropriate coefficients.

$x=\frac{-b}{2a}=\frac{-25}{2(-5)}= \frac{25}{10}=\frac{5}{2}$

Since the leading coefficient of is negative, the parabola opens down, which indicates that there will be a maximum.

The answer is: $\textup{Max at: }x=\frac{5}{2}$

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Question

Factorize:

Answer

To simplify , determine the factors of the first and last term.

The factor possibilities of :

The factor possibilities of :

Determine the signs. Since there is a positive ending term and a negative middle term, the signs of the binomials must be both negative. Write the pair of parenthesis.

These factors must be manipulated by trial and error to determine the middle term.

The correct selection is .

The answer is .

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Question

Find the location of the vertex of the parabola:

Answer

Multiply the two through the binomial.

Now that this is in the order of the polynomial , we can use the vertex formula.

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

Substitute the known coefficients.

$x=-\frac{b}{2a} =-\frac{6}{2(6)}=-\frac{1}{2}$

The answer is: $x=-\frac{1}{2}$

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Question

Find the location of the vertex for the parabola. Is it a max or min?

Answer

The polynomial is written in the form of:

This is the standard form for a parabola.

Write the vertex formula, and substitute the known values:

$x=-\frac{b}{2a}=-\frac{6}{2(-2)} = \frac{6}{4} =\frac{3}{2}$

The vertex is at: $x=\frac{3}{2}$

Since the coefficient of is negative, the curve will open downward, and will have a maximum.

The answer is: $\textup{Max at }x=\frac{3}{2}$

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Question

A particular parabola has it's vertex at , and an x-intercept at the origin. Determine the equation of the parabola.

Answer

General parabola equation:

Vertex formula:

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

Where is the value at the vertex.

Combining equations:

Plugging in values for vertex:

Solving for :

Returning to:

combining equations:

Plugging in values of given intercept:

Solving for

Plugging in value:

Plugging in values for the vertex:

Final equation:

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Question

Determine the vertex given the function:

Answer

The parabola is provided in the form of .

Notice that the variable in this equation is zero.

This means that:

$x=-\frac{b}{2a} = 0$

Substitute this value back into the original equation to determine the y-value.

The vertex is located at:

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Question

What is the point of the vertex of the parabola ? Is it a maximum or minimum?

Answer

It is not necessary to FOIL the binomial in order to solve for the vertex. Switch the terms of the quantity , and this equation will be in vertex form:

Set the inner quantity equal to zero.

Subtract three on both sides.

Divide by negative two on both sides.

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

The location of the vertex is at $x = \frac{3}{2}$ .

To determine the point, substitute the value $x = \frac{3}{2}$ back to the original equation.

$y=3(3-2(\frac{3}{2}))^2-1 = 3(3-3)^2-1 = 0-1 = -1$

The point of the vertex is at: $( \frac{3}{2},-1)$

Since this parabola opens upward, the point of the vertex will be a minimum.

The answer is: $( \frac{3}{2},-1), \textup{Minimum}$

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Question

Where is the vertex located for the given function?

Answer

Write the vertex formula.

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

The given equation is already in standard polynomial form.

Substitute the known values into the formula.

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

Substitute this value back into the original equation to determine the y value.

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

Simplify this expression.

$y=-2(\frac{1}{16})+\frac{1}{4}-1 = -\frac{1}{8}+\frac{1}{4}-1$

$y=-\frac{1}{8}+\frac{2}{8}-\frac{8}{8}=-\frac{7}{8}$

The vertex is located at: $(-\frac{1}{4},-\frac{7}{8})$

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Question

Which of these functions represent a parabola?

Answer

A parabola is a curve that is represented by a quadratic function. In this case, the only answer that qualifies is . The other answers represent straight lines, and other types of curves.

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Question

Where is the vertex located for ?

Answer

Rewrite the equation in standard polynomial form.

Write the vertex formula and substitute the known coefficients.

$x=-\frac{b}{2a} = -\frac{5}{2(-3)} = \frac{5}{6}$

The x-value of the vertex is $\frac{5}{6}$ .

Substitute this value back to the original equation.

$y=-3(\frac{5}{6})^2+5(\frac{5}{6})$

$y=-3(\frac{5}{6})(\frac{5}{6})+5(\frac{5}{6})$

$y=-\frac{25}{12}+\frac{25}{6} = -\frac{25}{12}+\frac{25(2)}{6(2)} =-\frac{25}{12}+\frac{50}{12}$

$y=\frac{25}{12}$

The vertex is located at: $(\frac{5}{6},\frac{25}{12} )$

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