Algebra II - Parabolic Functions

What are the -intercepts of the equation?

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

There are no -intercepts.

Explanation

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

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

Which of the following functions represents a parabola?

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

$f(x) = \frac{4x^{2}}{x^{3}}$

$f(x) = x^{2} - y^{2}$

Explanation

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.

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

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

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

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

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

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

Explanation

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:

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

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

None of these

Explanation

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:

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

$\frac{1}{2} \ and -\frac{3}{4}$

$-\frac{1}{2}\ and -\frac{3}{4}$

$-\frac{3}{2}\ and\ \frac{1}{4}$

$-\frac{3}{2}\ and\ -\frac{1}{4}$

$-\frac{1}{2}\ and\ \frac{3}{4}$

Explanation

Assume y=0,

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

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

Which of these functions represent a parabola?

Explanation

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.

Find the coordinates of the vertex of this quadratic function:

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

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

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

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

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

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

Explanation

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

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

Explanation

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

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

$\textup{Max at }x=\frac{3}{2}$

$\textup{Min at }x=\frac{3}{2}$

$\textup{Max at }x=\frac{2}{3}$

$\textup{Min at }x=\frac{2}{3}$

$\textup{Max at }y=\frac{3}{2}$

Explanation

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

Factorize:

Explanation

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

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 .

Parabolic Functions

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