SAT Math Lesson 3: Evaluating Functions, Linear Functions, and Quadratic Functions

Evaluating Functions

An understand of functions is necessary to do well on the SAT. Some questions will ask to evaluate functions. More complex questions will ask you to determine the independent value (normally the $x$ -value) that results in the dependent value (normally the $y$ -value)

Basic Functions

Let’s start by looking at the equation of a line, $y=mx+b$ .

We can find points on the line by substituting in values for $x$ and evaluating it to find the associated values of $y$ .

For example, if we have the equation $y=2x+1$ we can substitute in some values for $x$ and find the associated $y$ -values.

When $x=3$ , we would have $y=2(3)+1=7$ . The value of the function would be 7 when $x$ is 3.

We can find many different values for $y$ by substituting different values for $x$ . Here is a table showing different $x$ -values and different $y$ -values (function values).

x	2x+1	y
0	2(0)+1	1
1	2(1)+1	3
2	2(2)+1	5
3	2(3)+1	7
4	2(4)+1	9

We can plot those points on a graph

There are more values to the function than just the ones we evaluated. Every -value can be evaluated for this function. Here are some more:

x	2x+1	y
2.1	2(2.1)+1	5.2
2.2	2(2.2)+1	5.4
2.3	2(2.3)+1	5.6
2.4	2(2.4)+1	5.8
2.5	2(2.5)+1	6.0
2.6	2(2.6)+1	6.2

Let’s include those points on the graph also.

As you can see, the function is taking shape. We can see a straight line forming. Here is the graph of the function along with the plotted points. If all of the points were graphed for when equals 2.00001, 2.00002, etc. and even smaller values of , we would end up with a continuous function.

f(x), g(x), and h(x)

We looked at the simple function $y=2x-1$ . It can be written as $f(x)=2x-1$ , $g(x)=2x-1$ , $h(x)=2x-1$ , $p(x)=2x-1$ , $q(x)=2x-1$ , etc.

Don’t be confused when you see the notation $f(x)$ , it is similar to $y$ .
$f(x)$ is the notation used to refer to the function when it is in terms of $x$ .
However, if we want to show the value of a function at a specific $x$ -value, we would substitute the value in for $x$ .
For example
$f(x)=-2x-5$ ,
$f(3)=-2(3)-5$ ,
$f(3)=-11$
Let’s look at a table

If we want to find out what $f(3)$ or what the value of the function is when $x=3$ , we would look at the value of $f(x)$ when the $x$ -value is $3$ . In this case $f(3)=4$ .

If we want to find out what $f(7)$ or what the value of the function is when $x=7$ , we would look at the value of $f(x)$ when the $x$ -value is $7$ . In this case $f(7)=5$ .

Evaluating Functions from a Graph

A graph of a function, $f(x)$ , will likely have the vertical axis labeled $f(x)$ instead of $y$ . The horizontal axis will be labeled $x$ .

A graph of $s(t)$ will likely have the vertical axis labeled $s(t)$ instead of $y$ and the horizontal axis labeled $t$ instead of $x$ .
Here is an example.

We can see that the value of the graph changes at different spots on the graph.
The value of $f(x)$ is $2$ when $x$ is $-1$ .

$f(2)=4$ ,

$f(2.2)=4$ ,

$f(3)=4$ ,

and $f(4)=4$ .

In fact, the value of the function is $4$ when $x$ is equal to or greater than $2$ and less than or equal to $5$ .

Example

Based on the graph above, what is

a) $f(2)$ ?
b) The value of the function when $x$ is $3$ ?
c) The value of $x$ when $f(x)$ is $5$ ?

Explanation & Solution

a) $f(2)$ is the $y$ -value when $x=2$ . In the graph, when $x$ is 2, $y$ is 3.

b) The value of a function is the $y$ -value, so the value of the function when $x$ is 3 is the $y$ -value when $x$ is 3. In the graph, when $x=3$ , $y=2$ .

c) To find the value of $x$ when $f(x)$ is 5, we have to find the point (or points) on the graph where the $y$ -value is 5. They $y$ -value is 5 when $x$ is 1 and when $x$ is 3.75, so the answer could be 1 or 3.75.

Linear Functions

Linear functions are a big part of the SAT. Some questions can be direct and ask about properties of a line, while others may relate to a real-world problem in which something increases at a constant rate. On the calculator section (Section 4 of the SAT), in which there are data & probability questions, it’s likely there will be a question asking about the trendline of a scatterplot. The trendline will likely be a linear function.

Equation of a Line

The equation of a line is typically written in slope-intercept form, $y=mx+b$ .

It is called slope-intercept form because you can easily identify the slope and $y$ -intercept.

$y=mx+b$
where $m$ is the slope and $b$ is the $y$ -intercept.

Example
In the equation $y=2x+3$ , the slope is 2 and the $y$ -intercept is 3.
A slope of 2 really means the slope is $\dfrac{2}{1}$ . As you move from left to right, the line goes up 2 and right 1.
A $y$ -intercept of 3 means the slope intersects the $y$ -axis at 3. This is a good starting point when graphing.

Example
In the equation $y=-3/2 x+1$ , the slope is $=-\dfrac{3}{2}$ and the $y-intercept is 1. A slope of <img src='https://s0.wp.com/latex.php?latex&bg=ffffff&fg=000000&s=0' alt='' title='' class='latex' />=-\dfrac{3}{2}$ means as you move from left to right, the line goes down 3 and right 2.

A $y$ -intercept of 1 means the slope intersects the $y$ -axis at 1. This is a good starting point when graphing.

Parallel and Perpendicular Lines

Lines that are parallel have the same slope.

For example, if the equation of one line is $y=3x-9$ , a line parallel to it could have the equation
$y=3x+742$ . As long as the lines have the same slope and different $y$ -intercepts, they’re parallel.

Lines that are perpendicular, have slopes that are the negative reciprocal of each other.
For example, if the slope of one line is $\dfrac{2}{5}$ then the slope of a line perpendicular to it will be $-\dfrac{5}{2}$ .

Linear Functions

Linear functions can be written in a form similar to the equation of a line, $y=mx+b$ .

Linear functions are functions that increase or decrease at a constant rate, such as the amount of money paid for a gym membership. The membership may require an initial payment to join, followed by a monthly payment. The total amount of money paid increases at a constant rate since each month the total increases by the set monthly payment.

For example, it may cost $30 to join a gym plus $15 each month. After one month of membership you may have paid $45 ($30 to join and $15 monthly payment). The next month you’ll pay another $15, so the total paid will then be $60. The following month will be another $15, so the total paid will then be $75. Each month the total increases by $15. Let’s graph it.

Since we have a starting value of $30 and a rate of increase of 15, we can use the $y=15x+30$ to represent the function. Below is a graph including the function and points.

The key point is that if something increases at a constant rate, it can be modeled using the equation of a line as a linear function. The rate of change will be the slope and then initial value is typically the y-intercept.

Note about the variables.

Keep in mind that the variables used in a linear function do not have to be $x$ and $y$ . The variables can be any letter.

In the previous example, the total cost may be represented by $C$ and each month by $m$ . The function would be written as $C=15m+30$ . In this case m is the variable for number of months and 30 is the slope.

The function is sometimes also written as $C(m)$ instead of just $C$ . $C(m)$ is simply just a way of stating that the cost $C$ is dependent on the months, $m$ . In other words, $C$ is a function of $m$ .

Example
A company’s monthly expense is $20,000 each month for rent and other fixed costs. The products that the company creates cost $120 to produce. Write a function that represents the total cost $C$ in terms of the number of products produced, $p$ .

In this case, the cost increases at a constant rate. It increases by $120 for each product produced, so the slope is $120. Even if no products are produced, there is a cost of $20,000, so that is the starting value or $y$ -intercept.
$C=120p+20,000$

Quadratic Functions

Quadratic functions are a big part of the SAT on both sections: 3 and 4.

What are Quadratic Functions

Quadratic functions have the basic format of

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

where $a$ , $b$ and $c$ are constants.

It isn’t necessary to have a $bx$ term or a $c$ term, but it must have an $ax^2$ term.

Example of quadratics are

$f(x)=x^2+3x-5$ ,

$f(x)=2x^2$ ,

$f(x)=-3x^2-5$ ,

and $f(x)=x^2-5$ .

Graph of a Quadratic Functions

A graph of a quadratic function is a parabola (U-shaped).

The vertex is the point on the graph where the parabola changes from decreasing to increasing or from increasing to decreasing.

Parabolas are symmetric over the axis- of-symmetry, which passes through the vertex.

Zeros (Roots) of a Quadratic Function

The zeros, sometimes referred to as roots, of a function are the x-values that make the function (y-value) equal to zero. We can easily identify the zeros of a function from its graph. The graph below is zero (intersects the x-axis) at -1 and 3.

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Solving for Zeros by Factoring

To solve for the zeros of a function, set the function equal to zero and then solve.
Example
What are the zeros of the function f(x)=x^2-7x+6?

Explanation & Solution

Set the function equal to zero.

x^2-7x+6=0

Factor

Because the terms have to multiply to equal positive 6, the two terms must be the same sign (both positive or both negative). Since they have to add up to negative 7, they should both be negative.
-6 ×-1=6 and -6+ -1=-7, so we should use -6 and -1.

x^2-7x+6=0

(x-6)(x-1)=0

Now that we have the factors, we should set each one of the factors equal to zero and solve for x.

x-6=0 and x-1=0

So x=6 and x =1.

The zeros are 6 and 1. Let’s look at a graph of f(x)-7x+6 to see how it relates to the graph.

Solving for Zeros with the Quadratic Formula

You can always find the zeros of a quadratic function by using the quadratic formula.

$x=\dfrac{-b+-\sqrt{b^2-4ac}}{2x}$
If you can’t figure out the factors or if it isn’t factorable, use the quadratic formula.

Make sure to use parenthesis when you substitute the values in.

Let’s look at the same function, we looked at earlier $f(x)=x^2-7x+6$ .

Example

What are the zeros of the function $f(x)=x^2-7x+6$ ?

$f(x)=x^2-7x+6$

$a=1$ , $b=-7$ , $c=6$ ,

$x=\dfrac{-b+-\sqrt{b^2-4ac}}{2x}$ ,

$x=\dfrac{-(-7)+-\sqrt{(-7)^2-4(1)(6)}}{2(1)}$ ,

$x=\dfrac{7+-\sqrt{49-24}}{2}$ ,

$x=\dfrac{7+-\sqrt{25}}{2}$ ,

$x=\dfrac{7+-5}{2}$ ,

From this, we get two possible solutions

$x=\dfrac{7+5}{2}$ and $x=\dfrac{7-5}{2}$

which results in

$x=\dfrac{12}{2}$ and $x=\dfrac{2}{2}$

simplified to

$x=6$ and $x=1$

One Zero and No Zeros

The graph of a quadratic function has a U shape, so it’s possible that it will only touch the -axis once or even not at all, resulting in one zero or no zeros.

Graph of a quadratic function with one zero

The function only intersects the -axis once, so there is only one zero (root).

Graph of a quadratic function with no zero

The function doesn’t intersect the -axis, so there are no zeros (roots).

Example of One Zero

The function $f(x)=x^2-6x+9$ only has one zero. We can see by factoring or using the quadratic formula.

From factoring:

$x^2-6x+9=0$ ,

$(x-3)(x-3)=0$ ,

$x=3, x=3$

From the quadratic formula:

$x^2-6x+9=0$ ,

$x=\dfrac{-b+-\sqrt{b^2-4ac}}{2a}$ ,

$x=\dfrac{-(-6)+-\sqrt{(-6)^2-4(1)(9)}}{2a}$ ,

$x=\dfrac{6+-\sqrt{36-36}}{2}$ ,

$x=\dfrac{6+-0}{2}$

so, our solutions become

$x=\dfrac{6+0}{2}$ and $x=\dfrac{6-0}{2}$

but we end up with the same solution

$x=3$

Example of No Zeros

The function $f(x)=x^2-6x+10$ doesn’t have a zero.

We can’t factor it, so let’s look at it by using the quadratic formula.

$x=\dfrac{-b+-\sqrt{b^2-4ac}}{2a}$ ,

$x=\dfrac{-(-6)+-\sqrt{(-6)^2-4(1)(10)}}{2(1)}$ ,

$x=\dfrac{-(-6)+-\sqrt{36-40}}{2}$ ,

$x=\dfrac{-(-6)+-\sqrt{-4}}{2}$ ,

We end up with the square root of a negative number, which is not real, so there are no real roots.

Finding the Number of Zeros

Sometimes on the SAT, a question will ask for the number of roots or a similar question.

The following questions are all asking the same thing:

To find the number of real solutions to $x^2+3x+4=0$
To find the number of zeros the function $f(x)=x^2+3x+4$ has
To find number of roots of the function $f(x)=x^2+3x+4$ has

In each of the above questions, we would find the zeros using a graph, factoring, or the quadratic formula.

Sum and Product of Roots

To find the sum or product of the roots of a quadratic function, you could solve for the roots and then add them together to find the sum or multiply them to find the product.

The problem is that if the roots are irrational, such as $3+\sqrt{5}$ and $3-\sqrt{5}$ , it could be a little difficult to find the sum or product.

Instead, we can use two formulas to find the sum or product of the roots immediately.

If the function is in the form $f(x)=ax^2+bx+c$ , we can use the following formulas:

sum of roots= $\dfrac{-b}{a}$

product of roots= $\dfrac{c}{a}$

Example

Find the sum and product of the roots of the function $g(x)=2x^2+3x-8$ .

Explanation & Solution

sum of roots= $\dfrac{-b}{a}=\dfrac{-3}{2}$

product of roots= $\dfrac{c}{a}=\dfrac{-8}{2}=-4$

Forms of the Quadratic Function

Here are three ways of writing the same function:
$f(x)=x^2+2x-8$
$f(x)=(x+4)(x-2)$
$f(x)=(x+1)^2-9$

The top form is just a standard form. From $f(x)=x^2+2x-8$ , we can identify the $y$ -intercept, $-8$ .

The second form, $f(x)=(x+4)(x-2)$ lets us readily identify the zeros (roots) of the function, which are $-4$ and $2$ . Since it is a factored form of the function, we would set each factor to zero and solve for $x$ in each. .

The third form of the function, $f(x)=(x+1)^2-9$ , lets us readily identify the vertex and axis of symmetry. This function shows the function $x^2$ being translated left 1 unit and down 9 units, which would put the vertex at $(-1,-9)$ . The axis of symmetry also passes through the vertex, so the axis of symmetry is $x=-1$ .

Example

Which of the following formats of the function $f(x)=x^2+2x-3$ can be used to quickly identify the vertex?
A. $(x-1)^2+4$
B. $(x+3)(x-1)$
C. $(x+2)(x+1)$
D. $(x+1)^2 -4$

Explanation & Solution

We can eliminate choices B and C because the answers are not written in vertex form. They are written in a way that we can identify the zeros, but not the vertex.
We should multiply A and D out to see if we get the same functions as the one given.

$(x-1)^2+4$ ,
$(x-1)(x-1)+4$ ,
$x^2-x-x+1+4$ ,
$x^2-2x+5$
Choice A doesn’t work out, so we should try choice D.
$(x+1)^2-4$ ,
$(x+1)(x+1)-4$ ,
$x^2+x+x+1-4$ ,
$x^2+2x-3$
Choice D is the answer.

Tricks for Finding the Vertex

Trick #1

If the function is written in standard form, $y=ax^2+bx+c$ , we can quickly identify the axis of symmetry of a quadratic function with the following formula:
$x=\dfrac{-b}{2a}$
You might recognize this. It’s the first part of the quadratic formula, $x=\dfrac{-b+-\sqrt{b^2-4ac}}{2a}$
The axis of symmetry passes through the vertex, so the $x$ -coordinate of the vertex is the same as the axis of symmetry.
Example
Write the vertex form of the function $y=x^2+6x-5$ .

Explanation & Solution
$x=\dfrac{-b}{2a}$ ,
$x=\dfrac{-6}{2(1)}$ ,
$x=-3$
This means the $x$ -coordinate of the axis of symmetry is -3. We can substitute -3 in for $x$ in the function to find the $y$ -coordinate of the vertex.
$y=x^2+6x-5$ ,
$y=(-3)^2+6(-3)-5$ ,
$y=9-18-5$ ,
$y=-14$
The vertex form of the equation is $y=(x+3)^2-14$ .

Trick #2

If the function is written in a form that allows us to immediately find the zeros, then the second trick we can use is that the vertex and axis of symmetry exist exactly in the middle of the two zeros. You can find the axis of symmetry ( $x$ -coordinate of the vertex) by taking the average of the two numbers.

For example, if a function is $f(x)=x^2+2x-3$ , then the zeros of a function are -3 and 1. We can find the axis of symmetry and the $x$ -coordinate of the vertex by taking the average of the two numbers.

Average of the zeros= $\dfrac{-3+1}{2}=\dfrac{-2}{2}=-1$ .
The vertex is located at $x=-1$ and the axis of symmetry is $x=-1$ .

Since we know the $x$ -coordinate of the axis of symmetry, we can substitute it into the functions to find the y-coordinate.
$f(x)=x^2+2x-3$ ,
$f(-1)=(-1)^2+2(-1)-3$ ,
$f(-1)=1-2-3$ ,
$f(-1)=-4$ ,
The vertex is $(-1,-4)$ .

Below is a diagram to illustrate the concept.

Since the vertex is $(-1,-4)$ and coefficient in front of $x^2$ is $1$ , then vertex form of the function is $f(x)=(x+1)^2-4$ .

Projectile Motion

A question that appears on the SAT has to do with using a quadratic function to represent the height of a projectile with respect to time. Let’s look at an example. The height of a projectile fired off a cliff can be determined using the function $h(t)=-16t^2+128t+768$ , for $0 \leq t \leq 12$ . From what height was the projectile fired?

The projectile was fired from a height of 768 units, the value of the $y$ -intercept.

Table of Contents

Evaluating Functions

Basic Functions

f(x), g(x), and h(x)

Evaluating Functions from a Graph

Linear Functions

Equation of a Line

Parallel and Perpendicular Lines

Linear Functions

Quadratic Functions

What are Quadratic Functions

Graph of a Quadratic Functions

Zeros (Roots) of a Quadratic Function

Solving for Zeros by Factoring

Solving for Zeros with the Quadratic Formula

One Zero and No Zeros

Finding the Number of Zeros

Sum and Product of Roots

Forms of the Quadratic Function

Tricks for Finding the Vertex

Trick #1

Trick #2

Projectile Motion