\displaystyle x	\displaystyle =	\displaystyle -\dfrac{b}{2a}	Equation for the x-coordinate of the vertex
	\displaystyle =	\displaystyle -\dfrac{12}{2(3)}	Substitute a=3 and b=12
	\displaystyle =	\displaystyle -\dfrac{12}{6}	Evaluate the multiplication
\displaystyle x	\displaystyle =	\displaystyle -2	Evaluate the division

We can substitute the x-coordinate of the vertex into the original equation in order to find the y-coordinate of the vertex.

\displaystyle g(x)	\displaystyle =	\displaystyle 3x^2 + 12x - 15	Original function
	\displaystyle =	\displaystyle 3\left(-2\right)^2 + 12\left(-2\right) - 15	Substitute x=-2
	\displaystyle =	\displaystyle 3\left(4\right) + 12\left(-2\right) - 15	Evaluate the exponent
	\displaystyle =	\displaystyle 12 - 24 - 15	Evaluate the multiplication
	\displaystyle =	\displaystyle -27	Evaluate the subtraction

The coordinates of the vertex of g(x) are \left(-2,\,-27\right). We can confirm this by looking at the graph:

-6

-5

-4

-3

-2

-1

-25

-20

-15

-10

-5

We can also see here that the axis of symmetry is the line:

x=-\dfrac{b}{2a}

The axis of symmetry always passes through the vertex.

Examples

Example 1

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

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.

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).

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).

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

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.

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

\text{Time in seconds }(x)

-1

\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

\text{Time in seconds }(x)

-1

\text{Height in feet }(y)

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

\text{Time in seconds }(x)

-1

\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.

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 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

Now we need to convert to standard form:

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

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

Example 3

Write the standard form equation of the function shown on the graph.

-2

-4

-2

Worked Solution

Create a strategy

Use the vertex to first write the equation in vertex form, then use another point on the parabola to solve for a, and finally convert to standard form.

Apply the idea

Vertex form is y=a\left(x-h\right)^2+k. The vertex is \left(8, -2\right), this gives us: y=a\left(x-8\right)^2+(-2)or y=a\left(x-8\right)^2-2

We can then substitute in the point \left(4,\,6\right) and solve for a.

\displaystyle y	\displaystyle =	\displaystyle a\left(x-8\right)^2-2	Vertex form equation
\displaystyle 6	\displaystyle =	\displaystyle a\left(4-8\right)^2-2	Substitute in x=4 and y=6
\displaystyle 6	\displaystyle =	\displaystyle a\left(-4\right)^2-2	Evaluate the subtraction
\displaystyle 6	\displaystyle =	\displaystyle 16a - 2	Evaluate the exponent
\displaystyle 8	\displaystyle =	\displaystyle 16a	Add 2 to both sides
\displaystyle \dfrac{8}{16}	\displaystyle =	\displaystyle a	Divide by 16 on both sides
\displaystyle \dfrac{1}{2}	\displaystyle =	\displaystyle a	Simplify the fraction

Substituting a=\dfrac12 we get the equation: y=\dfrac{1}{2}\left(x-8\right)^2 - 2

Now we need to convert to standard form:

\displaystyle y	\displaystyle =	\displaystyle \dfrac{1}{2}\left(x-8\right)^2 - 2	Equation in vertex form
\displaystyle y	\displaystyle =	\displaystyle \dfrac{1}{2}\left(x^2 - 16x + 64\right) - 2	Expand the binomial
\displaystyle y	\displaystyle =	\displaystyle \dfrac{1}{2}x^2 - 8x + 32 - 2	Distributive property
\displaystyle y	\displaystyle =	\displaystyle \dfrac{1}{2}x^2 - 8x + 30	Combine like terms

y=\dfrac{1}{2}x^2 - 8x + 30 is the standard form equation of the function on the graph.

Reflect and check

Alternatively we could have used the zeros and factored form.

Using the zeros of 6 and 10, we can write the equation in factored form y=\left(x - x_1\right)\left(x - x_2 \right).This gives us:y=a(x - 6)(x - 10)

We can then substitute in the point \left(4,\,6\right) and solve for a.

\displaystyle y	\displaystyle =	\displaystyle a\left(x - 6\right)\left(x - 10\right)	Factored form equation
\displaystyle 6	\displaystyle =	\displaystyle a\left(4 - 6\right)\left(4 - 10\right)	Substitute x=4 \text{ and } y=6
\displaystyle 6	\displaystyle =	\displaystyle a\left(-2\right)\left(-6\right)	Evaluate the subtraction
\displaystyle 6	\displaystyle =	\displaystyle 12a	Evaluate the multiplication
\displaystyle \dfrac{6}{12}	\displaystyle =	\displaystyle a	Divide by 12 on both sides
\displaystyle \dfrac{1}{2}	\displaystyle =	\displaystyle a	Simplify the fraction

So now we have the equation: y=\dfrac{1}{2}\left(x-6\right)\left(x-10\right)

Now we need to convert to standard form:

\displaystyle y	\displaystyle =	\displaystyle \dfrac{1}{2}\left(x-6\right)\left(x-10\right)	Equation in factored form
\displaystyle y	\displaystyle =	\displaystyle \dfrac{1}{2}\left(x^2 - 10x - 6x + 60\right)	Multiply the binomials
\displaystyle y	\displaystyle =	\displaystyle \dfrac{1}{2}\left(x^2 - 16x + 60\right)	Combine like terms
\displaystyle y	\displaystyle =	\displaystyle \dfrac{1}{2}x^2 - 8x + 32	Distributive property

Example 4

The whale jumps out the water at 3 seconds and reenters the water after 6.5 seconds. The whale reaches a maximum height of 49 feet after 4.75 seconds. Determine the equation in standard form that models the whale’s jump.

Worked Solution

Create a strategy

If we think of the water level as the x-axis, then the moments where the whale exits and reenters the water would represent the x-intercepts. The maximum height is the vertex of the parabola formed by the whale's jump path.

Apply the idea

Using the x-intercepts of 3 and 6.5, we can create the following equation, where x represents the time in seconds and y represents the height of the jump in feet:

y=a(x-3)(x-6.5)

Next, we can substitute the values of the maximum point, which occurs at (4.75,\, 49), to find the value of a.

\displaystyle y	\displaystyle =	\displaystyle a\left(x-3\right)\left(x-6.5\right)	Equation for whale's path
\displaystyle 49	\displaystyle =	\displaystyle a(4.75-3)(4.75-6.5)	Substitute y=49 and x=4.75
\displaystyle 49	\displaystyle =	\displaystyle a(1.75)(-1.75)	Evaluate the subtraction
\displaystyle 49	\displaystyle =	\displaystyle -3.0625a	Evaluate the multiplication
\displaystyle -16	\displaystyle =	\displaystyle a	Division property of equality

Now, we know the factored form of the equation that models the whale's jump:

y=-16(x-3)(x-6.5)

The last step is to get it into standard form. We can do this by multiplying all the factors together.

\displaystyle y	\displaystyle =	\displaystyle -16\left(x-3\right)\left(x-6.5\right)	Equation for whale's path
\displaystyle y	\displaystyle =	\displaystyle -16(x^2-6.5x-3x+19.5)	Distributive property
	\displaystyle =	\displaystyle -16x^2+104x+48x-312	Distributive property
	\displaystyle =	\displaystyle -16x^2+152x-312	Combine like terms

The equation in standard form that models the whale's jump is y=-16x^2+152x-312.

Reflect and check

Notice that the parabola formed by the whale opens downward. If we did not have another point to help us find the value of a, our parabola would have been facing upward. Using the vertex helped us find a negative value for a which is what made the parabola face downward.

Idea summary

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}

The axis of symmtetry is also the x-coordinate of the vertex. To find the y-coordinate, you substitute the x-coordinate back into the original function. Therefore, the coordinates of the vertex are: \left(-\dfrac{b}{2a},\,f\left(-\dfrac{b}{2a}\right)\right)

Outcomes

A.F.2

The student will investigate, analyze, and compare characteristics of functions, including quadratic and exponential functions, and model quadratic and exponential relationships.

A.F.2b

Given an equation or graph, determine key characteristics of a quadratic function including x-intercepts (zeros), y-intercept, vertex (maximum or minimum), and domain and range (including when restricted by context); interpret key characteristics as related to contextual situations, where applicable.

A.F.2c

Graph a quadratic function, f(x), in two variables using a variety of strategies, including transformations f(x) + k and kf(x), where k is limited to rational values.

A.F.2d

Make connections between the algebraic (standard and factored forms) and graphical representation of a quadratic function.

A.F.2g

For any value, x, in the domain of f, determine f(x) of a quadratic or exponential function. Determine x given any value f(x) in the range of f of a quadratic function. Explain the meaning of x and f(x) in context.

7.04 Quadratic functions in standard form

Ideas

Quadratic functions in standard form

Exploration

Examples

Example 1

Example 2

Example 3

Example 4

Outcomes

A.F.2

A.F.2b

A.F.2c

A.F.2d

A.F.2g

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