SAT Math - Graphing Quadratic Functions

Consider the equation:

$y = 3x^{2} - 10x +5$

The vertex of this parabolic function would be located at:

$(\frac{5}{3}, \frac{-10}{3})$

$(\frac{-5}{3}, \frac{-10}{3})$

$(\frac{-5}{3}, 30)$

$(\frac{5}{3}, \frac{2}{25})$

$(\frac{5}{3}, \frac{27}{25})$

Explanation

For any parabola, the general equation is

$y = ax^{2} + bx +c$ , and the x-coordinate of its vertex is given by

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

For the given problem, the x-coordinate is

$x = \frac{-(-10)}{2(3)} = \frac{5}{3}$ .

To find the y-coordinate, plug $\frac{5}{3}$ into the original equation:

$y = 3(\frac{5}{3})^{2} - 10(\frac{5}{3}) +5 = \frac{-10}{3}$

Therefore the vertex is at $(\frac{5}{3}, \frac{-10}{3})$ .

Based on the figure below, which line depicts a quadratic function?

Red line

Blue line

Green line

Purple line

None of them

Explanation

A parabola is one example of a quadratic function, regardless of whether it points upwards or downwards.

The red line represents a quadratic function and will have a formula similar to $\small {f(x)=ax^2+bx+c}$ .

The blue line represents a linear function and will have a formula similar to $\small {f(x)=ax+b$ .

The green line represents an exponential function and will have a formula similar to $\small {f(x)=ax^b}$ .

The purple line represents an absolute value function and will have a formula similar to $\small {f(x)=\left |ax+b \right |}.$ .

Which of the following graphs matches the function ?

Graph

Graph1

Graph2

Graph3

Graph4

Explanation

Start by visualizing the graph associated with the function :

Graph5

Terms within the parentheses associated with the squared x-variable will shift the parabola horizontally, while terms outside of the parentheses will shift the parabola vertically. In the provided equation, 2 is located outside of the parentheses and is subtracted from the terms located within the parentheses; therefore, the parabola in the graph will shift down by 2 units. A simplified graph of looks like this:

Graph6

Remember that there is also a term within the parentheses. Within the parentheses, 1 is subtracted from the x-variable; thus, the parabola in the graph will shift to the right by 1 unit. As a result, the following graph matches the given function :

Graph

What is the vertex of the function $\small f(x)=2(x-4)^2+7$ ? Is it a maximum or minimum?

$\small (4,7)$ ; minimum

$\small (4,7)$ ; maximum

$\small (-4,7)$ ; minimum

$\small (-4,7)$ ; maximum

Explanation

The equation of a parabola can be written in vertex form: $\small f(x)=a(x-h)^2+k$ .

The point $\small (h,k)$ in this format is the vertex. If $\small a$ is a postive number the vertex is a minimum, and if $\small a$ is a negative number the vertex is a maximum.

$\small f(x)=2(x-4)^2+7$

In this example, $\small a=2$ . The positive value means the vertex is a minimum.

$\small h=4\ \text{and}\ k=7$

$\small (h,k)=(4,7)$

Which of the following parabolas is downward facing?

$y = -x^{2}$

$y = x^{2}$

$x = y^{2}$

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

$y = 5x^{2}+4$

Explanation

We can determine if a parabola is upward or downward facing by looking at the coefficient of the term. It will be downward facing if and only if this coefficient is negative. Be careful about the answer choice . Recall that this means that the entire value inside the parentheses will be squared. And, a negative times a negative yields a positive. Thus, this is equivalent to . Therefore, our answer has to be .

Give the -coordinate of the vertex of the parabola of the function

$f(x) = 2x^{2} - 6x + 5$

$\frac{1}{2}$

$\frac{3}{2}$

$\frac{37}{2}$

Explanation

The -coordinate of the vertex of a parabola of the form

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

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

Set :

$x= -\frac{-6}{2\cdot 2} = \frac{6}{4} = \frac{3}{2}$

The -coordinate is therefore $f\left ( \frac{3}{2} \right )$ :

$f(x) = 2x^{2} - 6x + 5$

$f \left ( \frac{3}{2} \right )= 2\left ( \frac{3}{2} \right )^{2} - 6\left ( \frac{3}{2} \right ) + 5$

$= 2\left ( \frac{9}{4} \right ) - 6\left ( \frac{3}{2} \right ) + 5$

$= \frac{9}{2} - 9 + 5 = \frac{1}{2}$ , which is the correct choice.

Give the -intercept(s) of the parabola of the equation

$f(x) = 2x^{2}+ 6x - 36$

and

$\left ( -\frac{3}{2}, 0 \right )$ and $\left ( \frac{3}{2}, 0 \right )$

$\left ( -36,0\right )$

and

The parabola has no -intercept.

Explanation

Set and solve for :

$f(x) = 2x^{2}+ 6x - 36$

$2x^{2}+ 6x - 36 = 0$

The terms have a GCF of 2, so

$2\left ( x^{2}+ 3x - 18 \right )= 0$

The trinomial in parentheses can be FOILed out by noting that and $6 \times (-3)= -18$ :

$2\left ( x+6\right ) (x-3)= 0$

Set each of the linear binomials to 0 and solve for :

$x + 6 = 0 \Rightarrow x = -6$

$x -3= 0 \Rightarrow x = 3$

The parabola has as its two intercepts the points and .

How many points of intersection could two distinct quadratic functions have?

, , and

only

and

Explanation

An intersection of two functions is a point they share in common. A diagram can show all the possible solutions:

Quadratics

Notice that:

and intersect times

and intersect time

and intersect times

The diagram shows that , , and are all possible.

Find the vertex form of the following quadratic equation:

$2x^{2} + 8x -1$

$2\left ( x +2 \right )^{2} - 9$

$2\left ( x + 2 \right )^{2} -5$

$\left ( x +2 \right )^{2} - 5$

$\left ( x - 2 \right )^{2} - 1$

$\left ( x + 2) \right ^{2} -1$

Explanation

Factor 2 as GCF from the first two terms giving us:

$2\left ( x^{2} \right + 4x ) -1$

Now we complete the square by adding 4 to the expression inside the parenthesis and subtracting 8 ( because $2\cdot 4=8$ ) resulting in the following equation: