High School Algebra I Study Guide for the STAAR test

Quadratic Functions and Equations

Quadratic functions and equations are 2nd degree functions and equations. Like linear functions and equations, you will need to be able to describe, write, and solve them. Eleven of the 54 questions on the test will assess these skills.

Writing

You will need to be able to write quadratic functions and equations in multiple ways.

Finding the Domain and Range

Here’s an example:

Find the domain and range for this graph of a quadratic function.

The domain is all real numbers or \(-\infty \lt x \lt \infty\), and the range is \(y\le6\). Note also that \(y=6\) is the maximum value for the graph.

Writing in Vertex and Standard Form

There are two forms for quadratic functions. They are:

Vertex form— \(f(x)=a(x-h)^2+k\), where \((h, k)\) is the vertex of the graph of the quadratic and \(a\) is some constant. As with linear functions, \(a\) is a vertical stretch of \(f\) if \(a\lt1\), and a vertical shrink if \(0\lt a \lt 1\). Also if \(a \lt 0\), then the graph of \(f\) is reflected about the \(x\)-axis.
Standard form— \(f(x)=ax^2+bx+c\), where \(a, b,\) and \(c\) are real numbers and \(a\ne0\).You may also need to put standard form into vertex form and visa-versa.

Example:

A parabola contains a point \((3,2)\) and a vertex of \((-1,4)\). Write its equation in standard form.

First, substituting the vertex, we get \(f(x)=a(x+1)^2+4\).

Now, using the given point, we can find \(a\) to get:

\[2=a(3+1)^2+4\] \[2=16a+4\] \[-2=16a\] \[a=-\frac{1}{8}\]

Our equation is now this, and by squaring and simplifying, we can put it in standard form as follows:

\[f(x)=-\frac{1}{8}(x+1)^2+4\] \[f(x)=-\frac{1}{8}(x^2+2x+1)+4\] \[f(x)=-\frac{1}{8}x^2-\frac{1}{4}x-\frac{1}{8}+4\] \[f(x)=-\frac{1}{8}x^2-\frac{1}{4}x+\frac{31}{8}\]

Using Solutions and Graphs

Also, for this test, you need to be able to write a quadratic function, given its solutions and graphs. A quadratic function in the form \(f(x)=a(x-x_1)(x-x_2)\) has real solutions of \(x_1\) and \(x_2\), where \(a\) is some constant that can be found, given any point on the graph of the quadratic function.

For example:

Write a quadratic function that has real solutions of \(-3\) and \(4\) and \(f(1)=5\) in standard form.

Substituting \(-3\) and \(4\) into form from above, we get \(f(x)=a(x+3)(x-4)\).

Now, substituting the given function value of \(f(1)=5\), we have:

\[5=a(1+3)(1-4)\] \[5=a(4)(-3)\] \[5=-12a\] \[a=-\frac{5}{12}\]

Now this one:

\[f(x)=-\frac{5}{12}(x+3)(x-4)\]

Foiling the binomials and simplifying our standard form is:

\[f(x)=-\frac{5}{12}(x+3)(x-4)\] \[f(x)=-\frac{5}{12}(x^2-x-12)\] \[f(x)=-\frac{5}{12}x^2+\frac{5}{12}x+5\]

Solving

You must be able to identify the equation of a quadratic function by looking at its graph and identify key attributes by looking at the graph or its equation. Additionally, you will find the zeros of a quadratic function using its factors.

Graph Attributes

When graphing a quadratic function, you will need to identify its key features, such as the \(x\)-intercepts(the zero of the function) and the \(y\)-intercept, the maximum or minimum value, and the axis of symmetry. Remember the axis of symmetry is a vertical line that cuts thru the \(x\) coordinate of the vertex of the graph.

Graph \(f(x)=-x^2+x-6\), using your technology and use the graph to state any \(x\)-intercepts(the zero of the function) and the \(y\)-intercept, the maximum or minimum value, and the axis of symmetry. The graph is shown here.

The zeros are \(x = -2\) and \(x = 3\).
The \(y\)-intercept is \(6\) and the vertex is approximately \((.5, 6.2)\) so the maximum value is approximately \(6.2\) and the axis of symmetry is approximately \(x = .5\).

Linear Factors and Zeros

You will use the linear factors to determine the zeros for a quadratic function. Remember the zeros are found by setting \(f(x) = 0\) and solving for \(x\) by setting each linear factor equal to zero, then solving each of them for \(x\).

For example:

What are the zeros for \(f(x)=-(x+3)(x-2)\)?

Setting \(f(x) = 0\), then setting the linear factors equal to zero, we get:

\[0=-(x+3)(x-2)\] \[x+3=0\;\text{or}\; x-2=0\] \[x=-3 \;\text{or}\; x=2\]

Transformations

Like linear functions, the transformed graph of the parent function \(f(x)=x^2\) can be found as follows:

For the function \(f(x)=x^2\), let \(g(x)=a \cdot (x-c)^2 +d\). The graph of \(g(x)\) is the graph of \(f(x)\), where:

\(a\) vertically stretches \(f\), if \(a \gt 1\), and vertically shrinks it, if \(0 \lt a \lt 1\). Also, if \(a \lt 0\), then the graph of \(f\) is reflected about the \(x\)-axis.
\(c\) horizontally shifts the graph. If \(c \gt 0\), the graph of \(f\) shifts to the right and shifts to the left if \(c \lt 0\).
\(d\) vertically shifts the graph. If \(d \gt 0\), the graph of \(f\) shifts upward, and shifts downward if \(d \lt 0\).

Discuss the transformations for the parent function \(f(x)=x^2\) in red, to obtain the transformed function \(y=g(x)\) in blue below.

Since the blue graph is reflected across the \(x\)-axis, then shifted \(1\) unit to the right and \(4\) units down (base this on the transformation of the vertex points), we have \(a=-1, c=1\), and \(d=-4\). Then, \(g(x) = -1\cdot(x-1)^2 +-4\) or \(g(x) = -(x-1)^2-4\)

Evaluating

Next, you will need to be able to evaluate quadratic functions and equations, including techniques to find their roots.

Methods for Solving

A quadratic equation in the form \(ax^2+bx+c=0\), where \(a \neq 0\), \(b\) and \(c\) are Real numbers, can be solved in the following ways:

Factoring, as was discussed earlier: If the quadratic factors, set the factors, which will be linear factors equal to zero, and solve each factor for the variable.
Use the square root property: If the quadratic takes on the form, \(ax^2-c=0\), where \(a \neq 0\) and \(c\) are real numbers, then the quadratic can be solved by using the square root as follows:

\[x= \pm\sqrt{\frac{c}{a}}\]

If the quadratic does not factor, you can always use the quadratic formula:

\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

If \(b^2-4ac \lt 0\), then the roots or solutions for the quadratic are imaginary numbers.

Graphing, as was done earlier, but sometimes only an approximate value can be found.

Here is an example:

Solve \(2x^2+5x-3=0\)

Factoring the quadratic into linear factors, we get:

\[2x^2+5x-3=0\] \[(2x-1)(x+3) = 0\] \[2x-1=0 \;\text{or}\; x+3=0\] \[x=\frac{1}{2}\; \text{or}\; x=-3\]

Find the zeros for \(x^2 +2x=4\)

First, set the quadratic equal to zero, then we have \(x^2+2x-4\)

You can try, but the quadratic will not factor, so, using the quadratic formula, where \(a=1, b=2\) and \(c=-4\), we get:

\[x=\frac{-2 \pm \sqrt{2^2-4(1)(-4)}}{2(1)}\] \[x=\frac{-2 \pm \sqrt{20}}{2}\] \[x=\frac{-2 \pm2 \sqrt{5}}{2}\] \[x=-1\pm \sqrt{5}\]

or we can say:

\[x=-1+\sqrt{5} \;\text{or}\; x=-1-\sqrt{5}\]

Best Fit

A projectile is launched from a pad that is \(6\) feet above ground. As the projectile moves up and eventually lands on the ground, the distance the projectile is from the ground, \(d\), is recorded in the table below. Assuming the distance the projectile is from the ground at time \(t\) in seconds forms a quadratic function, write an equation for it.

\[\begin{array}{|c|c|c|} \hline \text{t (sec.)} & \text{d (ft.)} \\ \hline \text{0} & \text{6} \\ \hline \text{2} & \text{198} \\ \hline \text{5} & \text{246} \\ \hline \text{8} & \text{6} \\ \hline \end{array}\]

Input the time in L1 and the distance in L2 by entering STAT then EDIT on your graphing calculator. Since we are told the data fits a quadratic function, enter STAT then over to CALC and enter 5:QuadReg. A summary of what you did will appear on the display. Scroll down to Calculate and the quadratic equation will appear.

The answer is \(y=-16x^2 +128x+6\) so, in function form, we have \(d(t)=-16t^2+128t+6\).