Ideas
- Quadratic functions
Quadratic functions
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 characteristics 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
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y
Standard form
f\left(x\right) = ax^2+bx+c
c is the y-intercept of the graph
x=-\dfrac{b}{2a} is the equation of the axis of symmetry
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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 characteristics 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 characteristics:
- x-intercepts
- y-intercept
- vertex (absolute maximum or minimum)
- axis of symmetry
- increasing, decreasing and constant intervals
- end-behavior
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 characteristics represent in the context.
Examples
Example 1
Consider the graph.
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a
State the coordinates of the vertex of the parabola.
Worked Solution
Create a strategy
The vertex lies on the axis of symmetry of the parabola, and is the maximum value in this case.
Apply the idea
The coordinates of the vertex are \left(4, 6\right).
b
Write the equation of the parabola in vertex form.
Worked Solution
Create a strategy
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 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.
Apply the idea
Substituting x=2, and y=2 into y = a \left(x - 4\right)^{2} + 6:
| \displaystyle 2 | \displaystyle = | \displaystyle a \left(2 - 4\right)^{2} + 6 | Substitute x=2, and y=2 |
| \displaystyle 2 | \displaystyle = | \displaystyle 4a + 6 | Evaluate the parentheses |
| \displaystyle -4 | \displaystyle = | \displaystyle 4a | Subtract 6 from both sides |
| \displaystyle -1 | \displaystyle = | \displaystyle a | Divide both sides by 4 |
The equation of the parabola in vertex form is y = -\left(x -4\right)^2 + 6
Reflect and check
Let's use our knowledge of transformations to check our equation.
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Because the graph is a parabola, we know the parent function is f(x)=x^2. Identifying transformations from the parent function, we see the graph has been:
Reflected over x-axis (flipped vertically)
Translated right 4 units (shifted horizontally)
Translated up 6 units (shifted vertically)
Vertex form of a quadratic equation is the same as the tranformation form, f\left(x\right)=a\left(x-h\right)^2+k. The reflection corresponds to a=-1, the horizontal translation corresponds to h=4, and the vertical translation corresponds to k=6. Therefore, our equation y=-\left(x-4\right)^2+6 is correct.
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.
Worked Solution
Create a strategy
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 zeros of the equation when written in factored form, so one approach to solving would be to write the equation in factored form.
Apply the idea
Writing the given equation in factored form, we get: 0=-16t\left(t-8\right)
To find the possible solution(s), we must set each factor equal to 0 and solve.
| \displaystyle -16t | \displaystyle = | \displaystyle 0 | Set the first factor equal to 0 |
| \displaystyle t | \displaystyle = | \displaystyle 0 | Divide both sides by -16 |
| \displaystyle t-8 | \displaystyle = | \displaystyle 0 | Set the second factor equal to 0 |
| \displaystyle t | \displaystyle = | \displaystyle 8 | Add 8 to both sides |
The solution t=0 represents 0 seconds after the ball was hit, which is the inital time. Because we know that the ball started on the ground, we can discount the solution of t=0.
This means the solution is t=8. The ball will hit the ground after 8 seconds.
Reflect and check
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h
We can check our solutions by graphing with technology. Recall that the solution(s) to an equation of the form f\left(x\right)=0 are the x-value(s) of the x-intercepts.
Here, we are using t for the independent variable rather than x. The graph intercepts the t-axis at t=0 and t=8 which corresponds to the solutions we found.
b
Find the greatest height the ball reaches above the ground.
Worked Solution
Create a strategy
Since the vertex of a parabola lies on the axis of symmetry, the greatest height the ball will reach is exactly halfway 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.
Apply the idea
| \displaystyle h | \displaystyle = | \displaystyle -16t^2+128t | State the given equation |
| \displaystyle h | \displaystyle = | \displaystyle -16\left(4\right)^2+128\left(4\right) | Substitute t=4 |
| \displaystyle h | \displaystyle = | \displaystyle 256 | Simplify |
The maximum height the ball reaches is 256 feet.
Reflect and check
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h
Another way to find the axis of symmetry is using x=-\dfrac{b}{2a} since the given equation is in standard form. Since a=-16 and b=128:\begin{aligned}x&=-\dfrac{128}{2\left(-16\right)}\\&=\dfrac{-128}{-32}\end{aligned} which gives us x=4. Then, we would still find the height by substituting x=4 into the equation.
We could also have rearranged the equation to be in vertex form:
| \displaystyle y | \displaystyle = | \displaystyle -16\left(t^2-8t\right) | Factor out -16 |
| \displaystyle y | \displaystyle = | \displaystyle -16\left(t^2-8t+16\right)+256 | Complete the square |
| \displaystyle y | \displaystyle = | \displaystyle -16\left(t-4\right)^2+256 | Factor the perfect square trinomial |
We 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.
Worked Solution
Create a strategy
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.
Apply the idea
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.
The domain is \left[0, 8\right].
Reflect and check
The domain can also be written using set notation, as shown: \{t\vert 0 \leq t \leq 8\}
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.
Worked Solution
Create a strategy
To identify the x-intercepts we can rewrite the equation in factored form.
Apply the idea
To rewrite the function in factored form, we want to first factor out the scale factor. This will give us:y=2\left(x^2+2x-15\right)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\left(x-3\right)\left(x+5\right)
b
Rewrite the quadratic equation in a form that allows us to identify the coordinates of the vertex.
Worked Solution
Create a strategy
To identify the vertex coordinates we can rewrite the equation in vertex form.
Apply the idea
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\left(x^2+2x-15\right) | Factor out the scale factor |
| \displaystyle y | \displaystyle = | \displaystyle 2\left(x^2+2x+1-1-15\right) | Add and subtract \left(\frac{2}{2}\right)^2=1 |
| \displaystyle y | \displaystyle = | \displaystyle 2\left(\left(x+1\right)^2-1-15\right) | Factor the perfect square |
| \displaystyle y | \displaystyle = | \displaystyle 2\left(x+1\right)^2-2-30 | Distributive property of multiplication |
| \displaystyle y | \displaystyle = | \displaystyle 2\left(x+1\right)^2-32 | Simplify |
So the vertex form of the quadratic function is:y=2\left(x+1\right)^2-32
c
Sketch the graph of the quadratic function, labeling the x- and y-intercepts, and the vertex.
Worked Solution
Create a strategy
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.
Apply the idea
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
Reflect and check
When sketching a quadratic function, we do not need to include a scale for the axes if we label all the characteristics, since we only need those points to determine the equation of the parabola.
Idea summary
Quadratic functions could be in the following forms.
Vertex form
\displaystyle f\left(x\right)=a\left(x-h\right)^2+k
\bm{\left(h,k\right)}
are the coordinates of the vertex (of the quadratic function)
Factored form
\displaystyle f\left(x\right)=a\left(x-x_1\right)\left(x-x_2\right)
\bm{x_1 \text{ and } x_2}
are the x-values of the
x-intercepts
Standard form
\displaystyle f\left(x\right)=ax^2+bx+c
\bm{c}
is the y-intercept
\bm{x=-\dfrac{b}{2a}}
is the equation of the axis of symmetry
