Topics
2. Quadratic Functions & Equations
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United States of AmericaFL
Grade 912

2.01 Graphing quadratic functions

Lesson
Practice
Lesson

Concept summary

When graphing parabolas and solving quadratic equations it is often useful to have the function written in a particular form, depending on the context and what key features we are interested in.

Vertex Form

f\left(x\right) = a\left(x - h\right)^2 + k

\left(h, k\right) are the coordinates of the vertex (of the quadratic function).

Factored form

f\left(x\right)=a\left(x-x_1\right)\left(x-x_2\right)

x_1 and x_2 are the x-values of the x-intercepts.

Standard Form

f\left(x\right) = ax^2+bx+c

c is the y-intercept of the graph

x=-\frac{b}{2a} is the equation of the axis of symmetry.

In all of the above forms the value of a is the scale factor of the quadratic function, and indicates the direction of opening of the graph. This means a \neq 0.

Worked examples

Example 1

Consider the graph.

-6
-4
-2
2
4
6
8
10
x
-10
-8
-6
-4
-2
2
4
6
y
a

State the coordinates of the vertex of the parabola.

Approach

The vertex lies on the axis of symmetry of the parabola, and is the maximum value in this case.

Solution

The coordinates of the vertex are \left(4, 6\right).

b

Write the equation of the parabola in vertex form.

Approach

We want to write the equation of the parabola in the form y = a \left(x - h\right)^{2} + k, and have already identified the vertex, so we know the equation will be y = a \left(x - 4\right)^{2} + 6, for some value of a.

To find a we can substitute in the values of any other point on the parabola into the equation and solve for a. The intercepts are not easily identifiable, but we can see the parabola passes through the point \left(2, 2\right), so we can use this point.

Solution

Substituting x=2, and y=2 into y = a \left(x - 4\right)^{2} + 6, gives us2 = a \left(2 - 4\right)^{2} + 6which we can solve to get a = -1.

So the equation of the parabola in vertex form is y = -\left(x -4\right)^2 + 6

Reflection

Due to the nature of this parabola, and its key features, vertex form is the easiest and most convenient form to write this in. We can see that the x-intercepts are not integer values and so are hard to determine. If asked to write this in standard form, it is easiest to start by finding the vertex form and then expanding and simplifying the result to get y=-x^2+8x-10.

Example 2

A golfball is hit into the air and its height h feet above the ground at time t seconds after being hit is given by h = - 16t^{2} + 128t.

a

Assuming the ball starts at a height of 0 feet, determine when it will hit the ground.

Approach

The ball will hit the ground when h=0, so we want to solve the equation 0=-16t^2+128t. The values of t that will solve this equation correspond with the roots of the equation when written in factored form, so one approach to solving would be to write the equation in factored form.

Solution

Writing the given equation in factored form, we get: h=-16\left(t-8\right)\left(t-0\right) We know that the ball started on the ground so we can discount the solution of t=0. This means the solution is t=8.

The ball will hit the ground after 8 seconds.

Reflection

We could have substituted in h=0 into the initial equation and solved for t:

\displaystyle h\displaystyle =\displaystyle -16t^2+128tStating the given equation
\displaystyle 0\displaystyle =\displaystyle -16t^2+128tSubstituting h=0
\displaystyle 16t^2\displaystyle =\displaystyle 128tAdding 16t^2
\displaystyle t\displaystyle =\displaystyle 8Dividing by 16t

Notice that the extraneous solution of t=0 is removed when we divide by 16t.

b

Find the greatest height the ball reaches above the ground.

Approach

Since the vertex of a parabola lies on the axis of symmetry, the greatest height the ball will reach is exactly half way between the two x-intercepts, when being hit and when hitting the ground. As we know, the ball hits the ground after 8 seconds, so it reaches its greatest height after exactly 4 seconds. To find this height we can substitute t=4 into the initial equation and solve for h.

Solution

\displaystyle h\displaystyle =\displaystyle -16t^2+128tStating the given equation
\displaystyle h\displaystyle =\displaystyle -16\left(4\right)^2+128\left(4\right)Substituting t=4
\displaystyle h\displaystyle =\displaystyle 256Simplifying

The maximum height the ball reaches is 256 feet.

Reflection

Since the given equation is in standard form, we could have found the axis of symmetry by substituting into -\dfrac{b}{2a}=-\dfrac{128}{(2)(-16)} which gives us 4.

We could also have rearranged the equation to be in vertex form y=-16\left(t-4\right)^2+256We can see from this that the vertex is at \left(4, 256\right).

c

Find the domain constraint for h, so it fits the restrictions of hitting the golfball. Give your answer using interval notation.

Approach

The domain is constrained by two things, the fact that time starts at t=0 and that the ball hits the ground after 8 seconds. After this time the quadratic equation will not model the height of the ball.

Solution

As the boundary times of t=0 and t=8 are included in the domain we will use square brackets, to indicate that they are included in the domain.

The domain is \left[0, 8\right].

Outcomes

MA.912.AR.3.8

Solve and graph mathematical and real-world problems that are modeled with quadratic functions. Interpret key features and determine constraints in terms of the context.

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