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3. Quadratic Functions & Equations
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United States of AmericaTennessee
Algebra 2

3.01 Graphing quadratic functions

Lesson
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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).

x
y
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.

x
y
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.

x
y

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. If a>0 then the parabola will open upwards, and if a<0 then the parabola opens downwards. This also means a \neq 0.

If we want to reveal different key features of a parabola, we can rewrite the quadratic function in different forms. This can be useful when sketching a graph of a quadratic function where we want to show all the key features:

  • x-intercepts
  • y-intercept
  • Vertex coordinates
  • Axis of symmetry

In addition to this, we can also use the context of a quadratic function to determine if there is an appropriate domain and range, or interpret what the key features represent in the context.

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].

Example 3

Sketch the graph of the quadratic function y=2x^2+4x-30, labeling the following key features:

  • x-intercepts
  • y-intercept
  • Vertex coordinates

Consider the quadratic function: y=2x^2+4x-30

a

Rewrite the quadratic equation in a form that allows us to identify the x-intercepts.

Approach

To identify the x-intercepts we can rewrite the equation in factored form.

Solution

To rewrite the function in factored form, we want to first factor out the scale factor. This will give us:y=2(x^2+2x-15)We then want to find two values that have a product of -15 and a sum of 2. If we check all the factor pairs of -15, we can find that -3 and 5 satisfy these requirements. So the factored form of the quadratic function is:y=2(x-3)(x+5)

b

Rewrite the quadratic equation in a form that allows us to identify the coordinates of the vertex.

Approach

To identify the vertex coordinates we can rewrite the equation in vertex form.

Solution

To rewrite the function in vertex form, we again want to factor out the scale factor, and then use the complete the square method.

\displaystyle y\displaystyle =\displaystyle 2(x^2+2x-15)Factor out the scale factor
\displaystyle y\displaystyle =\displaystyle 2(x^2+2x+1-1-15)Add and subtract \left(\frac{2}{2}\right)^2=1
\displaystyle y\displaystyle =\displaystyle 2((x+1)^2-1-15)Factor the perfect square
\displaystyle y\displaystyle =\displaystyle 2(x+1)^2-2-30Distributive property of multiplication
\displaystyle y\displaystyle =\displaystyle 2(x+1)^2-32Simplify

So the vertex form of the quadratic function is:y=2(x+1)^2-32

c

Sketch the graph of the quadratic function, labeling the x- and y-intercepts, and the vertex.

Approach

We can identify the y-intercept from the standard form.

We can identify the x-intercepts from the factored form.

And we can identify the coordinates of the vertex from the vertex form.

Solution

From the standard form of a quadratic equation, the y-intercept is \left(0,c\right). In this case, it will be \left(0,-30\right).

From the factored form of a quadratic equation, the x-intercepts are \left(x_1,0\right) and \left(x_2,0\right). In this case, they will be \left(3,0\right) and \left(-5,0\right).

From the vertex form of a quadratic equation, the vertex coordinates are \left(h,k\right). In this case, they are \left(-1,-32\right).

So we can sketch the quadratic function, labeling all the key features:

x
y

Reflection

When sketching a quadratic function, we do not need to include a scale for the axes if we label all the key features, since we only need those points to determine the equation of the parabola.

Outcomes

A2.N.Q.A.1

Use units as a way to understand real-world problems.*

A2.N.Q.A.1.A

Choose and interpret the scale and the origin in graphs and data displays.

A2.F.IF.B.4

Graph functions expressed algebraically and show key features of the graph by hand and using technology.*

A2.F.IF.B.5

Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function.*

A2.F.IF.B.5.A

Rewrite quadratic functions to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a real-world context.

A2.MP1

Make sense of problems and persevere in solving them.

A2.MP2

Reason abstractly and quantitatively.

A2.MP3

Construct viable arguments and critique the reasoning of others.

A2.MP4

Model with mathematics.

A2.MP6

Attend to precision.

A2.MP7

Look for and make use of structure.

A2.MP8

Look for and express regularity in repeated reasoning.

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