Investigating $y=ax^2+bx$

Equations such as $y=x^2+x$ are called quadratic equations because the highest power of the independent variable, $x$, is 2. In this lab you will learn about the shapes of the graphs and other properties of equations such as $y=x^2+x$.

Roots and axis of symmetry

Question 1. The roots, or zeros of a function are the $x$-coordinates of the points where its graph touches the $x$-axis. For example, for the function you are looking at to the left, $y=x^2-6x$, the roots are at 0 and 6. The axis of symmetry of this parabola is the line $x=3$. Use the sliders to change the values of $a$ and $b$ in the equation $y=ax^2+bx$ and answer the following questions.

What are the shapes of the red graphs you are looking at?

Complete the following table.

Equation$a$$b$rootsaxis of
symmetry
$y=x^2+2x$ 1 2 0−2
$y=x^2+4x$
$y=2x^2+8x$
$y=2x^2+4x$
Question 2. Look at the table you completed for question 1. Where is the axis of symmetry of a parabola compared to its roots?
Question 3. If the roots of a parabola are at 0 and 12, where is its axis of symmetry?
Where is the axis of symmetry of a parabola with roots at 0 and −12?

You can also find the roots of an equation without graphing, as in the following example. To find the roots and axis of symmetry of $y=x^2-3x$ without graphing, use factoring (or the distributive law), as follows.

The roots are where $y=0$, so we have
$x^2-3x=0$ which factors into:
$x(x-3)=0$. So, $x=0$ OR $x-3=0$ → $x=3$

Roots: The roots of $y=x^2-3x$ are 0 and 3.

Axis of symmetry: The axis of symmetry of a parabola passes midway between its roots, so the axis of symmetry of this parabola is the line $x=1.5$.

Question 4. Find the roots of $y=x^2-4x$ by factoring. &
Use these roots to find the axis of symmetry of $y=x^2-4x$.

Use the sliders and graph to the left to check your answers.

Question 5. Find the roots of $y=ax^2+bx$ by factoring.
[Hint: The factored form of $ax^2+bx$ is $x(ax+b)$, so you should solve for $x$ in $x(ax+b)=0$. That is, either $x=0$ or $ax+b=0$.] 		&
Find the axis of symmetry of $y=ax^2+bx$.

Question 6. Your results from question 1 have been copied into the table below. Complete this table.

Equation$a$$b$axis of
symmetry		$$-b/{2a}$$
$y=x^2+2x$ 1 2
$y=x^2+4x$ 1 4
$y=2x^2+8x$ 2 8
$y=2x^2+4x$ 2 4
What relationship do you see between the axis of symmetry and $$-b/{2a}$$?

The axis of symmetry of $y=ax^2+bx$ is the line $$x=-b/{2a}$$.

Below you can see the graph of $y=x^2-6x$. The axis of symmetry of this parabola is the line:

$$x = {-b}/{2a} = {-(-6)}/{2(1)} = 6/2 = 3$$

Vertex

We want to find the vertex of this parabola. The vertex is on the axis of symmetry, so its $x$-coordinate is 3. The vertex is also a point on the parabola, so it satisfies the equation for the parabola. This means that if you plug the $x$-coordinate of the vertex into the equation, you will get the $y$-coordinate.

Plugging 3 for $x$ into $y=x^2-6x$ gives $y=(3)^2-6(3)$ → $y=9-18$ → $y=-9$
So the vertex is at the point $(3,-9)$.

Question 7. Find the roots of $y=2x^2-4x$ by factoring. &
What is the axis of symmetry of this parabola?
Use the method described above to find the vertex of $y=2x^2-4x$.

Check your answers using the sliders and graph to the left.