We learned about different forms of quadratic functions in lesson 10.02 Quadratic functions in factored form and again in lesson 10.03 Quadratic functions in vertex form . The form of a quadratic function highlights certain features. We learn about the standard form of a quadratic function in this lesson.
Move the a, b, and c sliders to transform the graph.
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What happens to the graph as the value of a changes?
What happens to the graph as the value of b changes?
What happens to the graph as the value of c changes?
The standard form of a quadratic equation, where a,b, and c are real numbers is:
\displaystyle y=ax^2+bx+c
\bm{a}
scale factor
\bm{b}
linear coefficient
\bm{c}
y-value of the y-intercept
The standard form of a quadratic equation allows us to quickly identify the y-intercept and whether the parabola opens up or down.
The axis of symmetry is the line:
x=-\dfrac{b}{2a}
As the vertex lies on the axis of symmetry, this equation also shows us the x-coordinate of the vertex. We can substitute the x-coordinate of the vertex into the original equation in order to find the y-coordinate of the vertex, and therefore the coordinates of the vertex.
Examples
Example 1
For the quadratic function y=3x^2-6x+8:
a
Identify the axis of symmetry.
Worked Solution
Create a strategy
We will use the formula x=-\dfrac{b}{2a},so we need to identify the values of a and b from the equation.
a=3,b=-6
Apply the idea
\displaystyle x
\displaystyle =
\displaystyle -\dfrac{b}{2a}
Equation for axis of symmetry
\displaystyle x
\displaystyle =
\displaystyle -\dfrac{-6}{2\left(3\right)}
Substitute b=-6 and a=3
\displaystyle x
\displaystyle =
\displaystyle -\dfrac{-6}{6}
Evaluate the multiplication
\displaystyle x
\displaystyle =
\displaystyle -(-1)
Evaluate the division
\displaystyle x
\displaystyle =
\displaystyle 1
Evaluate the multiplication
The axis of symmetry is x=1.
b
State the coordinates of the vertex.
Worked Solution
Create a strategy
Once we have the x-coordinate of the vertex from the axis of symmetry, we can substitute it into y=3x^2-6x+8 and evaluate to get y.
From part (a), we know that the axis of symmetry is x=1 so the x-coordinate of the vertex is x=1.
Apply the idea
\displaystyle y
\displaystyle =
\displaystyle 3x^2-6x+8
Given equation
\displaystyle y
\displaystyle =
\displaystyle 3\left(1\right)^2-6\left(1\right)+8
Substitute x=1
\displaystyle y
\displaystyle =
\displaystyle 3\left(1\right)-6\left(1\right)+8
Evaluate the exponent
\displaystyle y
\displaystyle =
\displaystyle 3-6+8
Evaluate the multiplication
\displaystyle y
\displaystyle =
\displaystyle 5
Evaluate the subtraction and addition
The vertex is \left(1,5\right).
c
State the coordinates of the y-intercept.
Worked Solution
Create a strategy
Since the y-intercept occurs when x=0, substitute x=0 into the equation and evaluate y.
When we are given an equation in standard form, the y-value of the y-intercept will be y=c.
Apply the idea
In this case, the value of c in the equation is 8.
So we have that the coordinates of the y-intercept are \left(0, 8\right).
d
Draw a graph of the corresponding parabola.
Worked Solution
Create a strategy
We have all the key features we need to create a graph. For more accuracy, we can use the axis of symmetry and y-intercept to find another point. This point will be a reflection of the y-intercept across the axis of symmetry.
We know that the parabola will open upwards because a>0.
Apply the idea
-1
1
2
3
x
2
4
6
8
y
Axis of symmetry: x=1
Vertex: \left(1,5\right)
y-intercept: \left(0,8\right)
Another point: \left(2,8\right)
Reflect and check
From the graph we can identify that the vertex form of the equation would be:
y=3\left(x-1\right)^2+5
Example 2
Naomi is playing a game of Kapucha Toli, where to start a play, a ball is thrown into the air. Naomi throws a ball into the air from a height of 6 feet, and the maximum height the ball reaches is 12.25 feet after 1.25 seconds.
a
Sketch a graph to model the height of the ball over time.
Worked Solution
Create a strategy
To sketch a graph, we'll use key points found by using the given information. We'll also use the units which are given, being feet and seconds.
We will let x represent the time since the ball was tossed in seconds.
We will let y represent the height of the ball in feet.
Apply the idea
It's given that the ball is thrown from a height of 6 feet. This means that at 0 seconds, the height of the ball is 6 feet. So our y-intercept is \left(0, 6 \right).
We're told the maximum height of the ball is at 12.25 feet after 1.25 seconds. The maximum height will occur at the vertex of the graph, so the vertex is \left(1.25, 12.25\right). This also means that our axis of symmetry is x=1.25.
We can use the axis of symmetry to determine a second point on the graph, the point across the axis of symmetry from the y-intercept. The point is \left(2.5, 6 \right).
We can sketch our graph by plotting the y-intercept, the vertex, and the point found with our axis of symmetry.
Now we need to identify an appropriate scale.
Kapucha Toli
1
2
3
\text{Time in seconds }(x)
-1
1
2
3
4
5
6
7
8
9
10
11
12
13
\text{Height in feet }(y)
We know that our graph will not go above y=12.25 feet and that any part of the graph that goes below the x-axis will not be viable, so graphing -1 \leq y \leq 13 going up by 1 will show the full picture.
We know that time starts at x=0 and the ball is on the way back down at x=2.5, so graphing 0 \leq x \leq 4 going up by 1 or 0.5 should be sufficient.
This is an appropriate way to label the axes.
Now we can graph the height of the ball over time.
Kapucha Toli
1
2
3
\text{Time in seconds }(x)
-1
1
2
3
4
5
6
7
8
9
10
11
12
13
\text{Height in feet }(y)
b
Predict when the ball will be 3 feet above the ground.
Worked Solution
Create a strategy
We can use the sketch of our graph to predict when the ball will be at 3 feet.
Apply the idea
We can draw a horizontal line from y=3 across until we reach the graph. After that, we can draw vertical line until we reach the x-axis to determine after how many seconds the ball is at 3 feet.
Kapucha Toli
1
2
3
\text{Time in seconds }(x)
-1
1
2
3
4
5
6
7
8
9
10
11
12
13
\text{Height in feet }(y)
We hit the x-axis around x=2.7. Therefore, the ball is 3 \text{ ft} above the ground after about 2.7 seconds.
Reflect and check
When reading from a graph, we often have to estimate. Any prediction between 2.6 and 2.9 would be reasonable in this case.
c
Write a quadratic equation in standard form to model the situation.
Worked Solution
Create a strategy
To write the equation, we can use the key points and the graph we've sketched in previous parts.
Apply the idea
Since we know 3 points, we can use the standard form and substitution in order to solve for a, b, and c for our standard form quadratic equation which is of the form y=ax^{2}+bx+c.
We know that c represents the y-value of the y-intercept which is \left(0, 6 \right):y=ax^2+bx+6
Now we can substitute our other two points to solve for a and b.
Next we can substitute in \left(1.25, 12.25\right).
\displaystyle y
\displaystyle =
\displaystyle ax^{2}+bx+6
Standard form of a quadratic with c=6
\displaystyle 12.25
\displaystyle =
\displaystyle a (1.25) ^{2}+b (1.25) +6
Substitute \left(1.25,12.25\right)
\displaystyle 12.25
\displaystyle =
\displaystyle 1.5625a+ 1.25b +6
Evaluate the exponent
\displaystyle 6.25
\displaystyle =
\displaystyle 1.5625a + 1.25b
Subtract 6 from both sides
Since we have two unknowns, we'll have to use our final point to create a second equation.
We'll now substitute in \left(2.5, 6 \right).
\displaystyle y
\displaystyle =
\displaystyle ax^{2}+bx+6
Standard form of a quadratic with c=6
\displaystyle 6
\displaystyle =
\displaystyle a (2.5) ^{2}+b (2.5) +6
Substitute \left(2.5,6\right)
\displaystyle 6
\displaystyle =
\displaystyle 6.25a + 2.5b +6
Evaluate the exponent
\displaystyle 0
\displaystyle =
\displaystyle 6.25a + 2.5b
Subtract 6 from both sides
Now we have two equations with two unknowns. We can solve this system using the substitution method. Let's first isolate for b in our second equation.
\displaystyle 0
\displaystyle =
\displaystyle 6.25a + 2.5b
Second equation
\displaystyle -6.25a
\displaystyle =
\displaystyle 2.5b
Subtract 6.25a from both sides
\displaystyle \frac{-6.25a}{2.5}
\displaystyle =
\displaystyle b
Divide by 2.5 on both sides
\displaystyle -2.5a
\displaystyle =
\displaystyle b
Evaluate the division
Now we can use the this in our first equation, letting b = -2.5 a.
\displaystyle 6.25
\displaystyle =
\displaystyle 1.5625a + 1.25b
First equation
\displaystyle 6.25
\displaystyle =
\displaystyle 1.5625a + 1.25 (-2.5a)
Substitute b = -2.5 a
\displaystyle 6.25
\displaystyle =
\displaystyle 1.5625a -3.125a
Evaluate the multiplication
\displaystyle 6.25
\displaystyle =
\displaystyle -1.5625a
Combine like terms
\displaystyle -4
\displaystyle =
\displaystyle a
Divide both sides by -1.5625
Therefore a=-4. Finally, we can use b = -2.5 a to solve for b.
\displaystyle b
\displaystyle =
\displaystyle -2.5a
\displaystyle b
\displaystyle =
\displaystyle -2.5(-4)
Substitute a = -4
\displaystyle b
\displaystyle =
\displaystyle 10
Evaluate the multiplication
Therefore b=10. Now it's time to piece it all together. Since a=-4, b=10, and c=6, we know that our equation in standard form is: y = -4x^{2} + 10 x + 6
Reflect and check
An alternative and simpler solution is to use the vertex to write it in vertex form and then using the intercept to solve for a.
Vertex form is y=a\left(x-h\right)^2+k. The vertex is \left(1.25, 12.25\right), this gives us: y=a\left(x-1.25\right)^2+12.25
We can then substitute in the point \left(0,6\right) and solve for a.
\displaystyle y
\displaystyle =
\displaystyle a\left(x-1.25\right)^2+12.25
Vertex form of a quadratic with vertex \left(1.25, 12.25\right)
\displaystyle 6
\displaystyle =
\displaystyle a\left(0-1.25\right)^2+12.25
Substitute in \left(0,6\right)
\displaystyle 6
\displaystyle =
\displaystyle 1.5625a+12.25
Evaluate the parentheses
\displaystyle -6.25
\displaystyle =
\displaystyle 1.5625a
Subtract 12.25 from both sides
\displaystyle \dfrac{-6.25}{1.5625}
\displaystyle =
\displaystyle a
Divide by 1.5625 on both sides
\displaystyle -4
\displaystyle =
\displaystyle a
Evaluate the division
So now we have the equation: y=-4\left(x-1.25\right)^2+12.25
The standard form of a quadratic equation highlights the y-intercept of a quadratic function.
\displaystyle y=ax^2+bx+c
\bm{a}
scale factor
\bm{b}
linear coefficient
\bm{c}
y-value of the y-intercept
The axis of symmetry is the line:
x=-\dfrac{b}{2a}
Outcomes
A.CED.A.2
Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.
F.IF.B.4
For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.
F.IF.C.7
Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases.
F.IF.C.7.A
Graph linear and quadratic functions and show intercepts, maxima, and minima.
F.IF.C.8
Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function.
F.BF.A.1
Write a function that describes a relationship between two quantities.
F.BF.A.1.A
Determine an explicit expression, a recursive process, or steps for calculation from a context.
