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7. Quadratic Functions
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United States of AmericaTennessee
Algebra 1

7.05 Quadratic functions in standard form

Lesson
Practice
Lesson

Concept summary

The standard form of a quadratic equation allows us to quickly identify the y-intercept and whether the parabola opens up or down.

The standard form of a quadratic equation is:

\displaystyle y=ax^2+bx+c
\bm{a}
The scale factor which tells us about the shape of the graph
\bm{b}
Helps us to find the axis of symmetry and vertex
\bm{c}
The y-value of the y-intercept

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.

Worked examples

Example 1

For the quadratic function y=3x^2-6x+8:

a

Identify the axis of symmetry.

Approach

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

Solution

\displaystyle x\displaystyle =\displaystyle -\dfrac{b}{2a}Equation for axis of symmetry
\displaystyle x\displaystyle =\displaystyle -\dfrac{-6}{2\left(3\right)}Substitute in the values
\displaystyle x\displaystyle =\displaystyle -\dfrac{-6}{6}Evaluate the denominator
\displaystyle x\displaystyle =\displaystyle -(-1)Evaluate the division
\displaystyle x\displaystyle =\displaystyle 1Evaluate the double negative

The axis of symmetry is x=1.

b

State the coordinates of the vertex.

Approach

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.

Solution

\displaystyle y\displaystyle =\displaystyle 3x^2-6x+8Given equation
\displaystyle y\displaystyle =\displaystyle 3\left(1\right)^2-6\left(1\right)+8Substitute x=1
\displaystyle y\displaystyle =\displaystyle 3\left(1\right)-6\left(1\right)+8Evaluate the exponent
\displaystyle y\displaystyle =\displaystyle 3-6+8Evaluate the multiplication
\displaystyle y\displaystyle =\displaystyle 5Evaluate the subtraction and addition

The vertex is \left(1,5\right).

c

State the coordinates of the y-intercept.

Approach

Since the y-intercept occurs when x=0, we can 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.

Solution

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.

Approach

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.

Solution

-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)

Reflection

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

For the quadratic function y = -x^2 + 7x - 10:

a

State the coordinates of the y-intercept.

Approach

Since the y-intercept occurs when x=0, we can substitute x=0 into the equation and determine the value of y.

When we are given an equation in standard form, the y-value of the y-intercept will be y=c.

Solution

In this case, the value of c in the equation is -10.

So we have that the coordinates of the y-intercept are \left(0, -10\right).

b

State the coordinates of the x-intercept(s).

Approach

Since the x-intercept(s) occur when y = 0, we can substitute y = 0 into the equation and solve for the values of x.

Since the equation is given in standard form, this will involve factoring the quadratic.

Solution

Setting y = 0, we have the equation -x^2 + 7x - 10 = 0. We can now solve this for x as follows:

\displaystyle -x^2 + 7x - 10\displaystyle =\displaystyle 0Substitute y = 0
\displaystyle x^2 - 7x + 10\displaystyle =\displaystyle 0Multiplication property of equality
\displaystyle \left(x - 2\right)\left(x - 5\right)\displaystyle =\displaystyle 0Factor the quadratic

The two values of x which satisfy this equation are x = 2 and x = 5. This tells us that the coordinates of the two x-intercepts are \left(2, 0\right) and \left(5, 0\right).

c

Sketch a graph of the corresponding parabola.

Approach

We now know the coordinates of the x- and y-intercepts of the function. We can use these to plot the parabola. For more accuracy, we can also note that the axis of symmetry will occur exactly halfway between the x-intercepts - so in this case, the axis of symmetry will be the line x = \frac{2 + 5}{2} = 3.5.

We also know that the parabola will open downwards because a<0.

Solution

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

y-intercept: \left(0, -10\right)

x-intercepts: \left(2,0\right) and \left(5, 0\right)

Example 3

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. Make sure to label the axes with an appropriate scale and units.

Approach

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.

Solution

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.

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.

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.

Approach

We can use the sketch of our graph to predict when the ball will be at 3 feet.

Solution

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 \text{ feet}

1
2
3
x \text{(sec)}
-1
1
2
3
4
5
6
7
8
9
10
11
12
13
y \text{(ft)}

We hit the x-axis around x=2.7. Therefore the ball is 3 \text{ feet} above the ground after about 2.7 \text{ seconds}.

Reflection

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.

Approach

To write the equation, we can use the key points and the graph we've sketched in previous parts.

Solution

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 represent the y-value of the y-intercept which is \left(0, 6 \right), so far get: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+6Standard form of a quadratic with c=6
\displaystyle 12.25\displaystyle =\displaystyle a (1.25) ^{2}+b (1.25) +6Substitute \left(1.25,12.25\right) into the equation
\displaystyle 12.25\displaystyle =\displaystyle 1.5625a+ 1.25b +6Evaluate the exponent
\displaystyle 6.25\displaystyle =\displaystyle 1.5625a + 1.25bSubtract 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+6Standard form of a quadratic with c=6
\displaystyle 6\displaystyle =\displaystyle a (2.5) ^{2}+b (2.5) +6Substitute \left(2.5,6\right) into the equation
\displaystyle 6\displaystyle =\displaystyle 6.25a + 2.5b +6Evaluate the exponent
\displaystyle 0\displaystyle =\displaystyle 6.25a + 2.5bSubtract 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 + 2.5bSecond equation
\displaystyle -6.25a\displaystyle =\displaystyle 2.5bSubtract 6.25a from both sides
\displaystyle \frac{-6.25a}{2.5}\displaystyle =\displaystyle bDivide by 2.5 on both sides
\displaystyle -2.5a\displaystyle =\displaystyle bSimplify

Now we can use the this in our first equation, letting b = -2.5 a.

\displaystyle 6.25\displaystyle =\displaystyle 1.5625a + 1.25bFirst equation
\displaystyle 6.25\displaystyle =\displaystyle 1.5625a + 1.25 (-2.5a)Substitute b = -2.5 a from second equation
\displaystyle 6.25\displaystyle =\displaystyle 1.5625a -3.125aEvaluate 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 10Evaluate 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

Reflection

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+12.25Using vertex form
\displaystyle 6\displaystyle =\displaystyle a\left(0-1.25\right)^2+12.25Substitute in \left(0,6\right)
\displaystyle 6\displaystyle =\displaystyle 1.5625a+12.25Evaluate the parentheses
\displaystyle -6.25\displaystyle =\displaystyle 1.5625aSubtraction property of equality
\displaystyle \dfrac{-6.25}{1.5625}\displaystyle =\displaystyle aDivision property of equality
\displaystyle -4\displaystyle =\displaystyle aDivision property of equality

So now we have the equation: y=-4\left(x-1.25\right)^2+12.25

Now we need to convert to standard form:

\displaystyle y\displaystyle =\displaystyle -4\left(x-1.25\right)^2+12.25Equation in vertex form
\displaystyle y\displaystyle =\displaystyle -4\left(x^2-2.5x+1.5625\right)^2+12.25Using the perfect square trinomial pattern
\displaystyle y\displaystyle =\displaystyle -4x^2+10x-6.25+12.25Distributive property
\displaystyle y\displaystyle =\displaystyle -4x^2+10x+6Combine like terms

We get the same answer of: y=-4x^2+10x+6.

Outcomes

A1.N.Q.A.1

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

A1.N.Q.A.1.A

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

A1.A.CED.A.2

Create equations in two variables to represent relationships between quantities and use them to solve problems in a real-world context. Graph equations with two variables on coordinate axes with labels and scales, and use the graphs to make predictions.*

A1.A.CED.A.3

Create individual and systems of equations and/or inequalities to represent constraints in a contextual situation, and interpret solutions as viable or non-viable.*

A1.A.REI.D.5

Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line).

A1.A.REI.D.6

Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x). Find approximate solutions by graphing the functions or making a table of values, using technology when appropriate.*

A1.F.IF.C.7

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

A1.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. *

A1.F.IF.C.8.A

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

A1.F.IF.C.9

Compare properties of functions represented algebraically, graphically, numerically in tables, or by verbal descriptions.*

A1.MP1

Make sense of problems and persevere in solving them.

A1.MP2

Reason abstractly and quantitatively.

A1.MP3

Construct viable arguments and critique the reasoning of others.

A1.MP4

Model with mathematics.

A1.MP6

Attend to precision.

A1.MP7

Look for and make use of structure.

A1.MP8

Look for and express regularity in repeated reasoning.

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