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Interpreting Quadratic Functions in Standard Form

There are several ways to write the rule of a quadratic function. Each form highlights certain characteristics of the parabola. Standard form is expressed as follows.

y=ax2+bx+cy=ax^2+bx+c

Here, a,a, b,b, and cc can be any real number, and a0.a\neq 0.
Concept

Characteristics of Quadratic Functions in Standard Form

When a quadratic function is written in standard form, it's possible to use a,a, b,b, and cc to determine characteristics of its graph. direction:upward when a>0,:downward when a<0y-intercept:(0,c)axis of symmetry:x=-b2a\begin{aligned} \textbf{direction} &: \text{upward when } a>0, \\ &\phantom{:} \text{downward when } a<0 \\ \mathbf{y} \textbf{-intercept} &: (0,c) \\ \textbf{axis of symmetry} &: x=\text{-}\dfrac{b}{2a} \end{aligned}

Concept

Direction

The direction of the graph is determined by the sign of a.a. To understand why, consider the quadratic function y=ax2. y=ax^2. Since all squares are positive, x2x^2 will always be positive. When aa is positive, then ax2ax^2 is also positive. Thus, when moving away from the origin in either direction, the graph extends upward. Similarly, when aa is negative, ax2ax^2 will be negative. Thus, the graph will extend downward for all xx-values.

Concept

yy-intercept

The yy-intercept of a quadratic function is given by c,c, specifically at (0,c).(0,c). This is because substituting x=0x=0 into standard form yields the following. y=ax2+bx+cy=a02+b0+cy=0+0+cy=c\begin{aligned} y&=ax^2+bx+c\\ y&=a \cdot {\color{#0000FF}{0}}^2 + b \cdot {\color{#0000FF}{0}} + c \\ y&= 0+0+c \\ y&=c \end{aligned}

Concept

Axis of Symmetry

The equation of the axis of symmetry can be found using the coefficients aa and b.b. It is derived from the fact that the axis of symmetry divides the parabola in two mirror images. Two points with the same yy-value are, thus, equidistant from the axis of symmetry. This gives rise to a quadratic equation where the solution is the axis of symmetry.
Concept

Symmetry of a Parabola

For all quadratic functions, the axis of symmetry will always intersect the parabola at its vertex. Additionally, two points with the same yy-coordinate will always be equidistant from the axis of symmetry. Move the three points to see how a parabola that passes through them looks.

Draw the graph

Exercise

The function f(x)=3x0.5x2f(x)=3x-0.5x^2 describes the height of the mouth of a tunnel. Here, xx is the distance from the lower left corner, and both xx and f(x)f(x) are in meters. Complete the table of values to graph the function and determine the width and height of the tunnel.

xx yy
00
11
22
33
44
55
66
Solution
To begin, we can complete the table of values by substituting the given xx-values into the function rule. We'll start with x=0.x=0.
f(x)=3x0.5x2f(x)=3x-0.5x^2
f(0)=300.502f({\color{#0000FF}{0}})=3\cdot {\color{#0000FF}{0}}-0.5\cdot {\color{#0000FF}{0}}^2
f(0)=0f(0)=0
We can perform the same process for the other xx-values.
xx 3x0.5x23x-0.5x^2 f(x)f(x)
1{\color{#0000FF}{1}} 310.5123\cdot {\color{#0000FF}{1}}-0.5\cdot {\color{#0000FF}{1}}^2 2.52.5
2{\color{#0000FF}{2}} 320.5223\cdot {\color{#0000FF}{2}}-0.5\cdot {\color{#0000FF}{2}}^2 44
3{\color{#0000FF}{3}} 330.5323\cdot {\color{#0000FF}{3}}-0.5\cdot {\color{#0000FF}{3}}^2 4.54.5
4{\color{#0000FF}{4}} 340.5423\cdot {\color{#0000FF}{4}}-0.5\cdot {\color{#0000FF}{4}}^2 44
5{\color{#0000FF}{5}} 350.5523\cdot {\color{#0000FF}{5}}-0.5\cdot {\color{#0000FF}{5}}^2 2.52.5
6{\color{#0000FF}{6}} 350.5623\cdot {\color{#0000FF}{5}}-0.5\cdot {\color{#0000FF}{6}}^2 00

To graph the function, we can plot the points and connect them with a smooth curve.

The graph gives an approximation of the height and width of the tunnel. We can think of the xx-axis as the ground. Thus, the distance between the xx-axis and the vertex, which is a maximum, gives the height of the tunnel, and the distance between the zeros gives the width of the tunnel.

From this graph, we can see that the height appears to be 4.54.5 meters and that the width appears to be 66 units. Because the zeros were found algebraically to be (0,0)and(6,0), (0,0) \quad \text{and} \quad (6,0), we can definitively state that the width of the tunnel is 66 meters. What remains is to algebraically determine the coordinate of the maximum (vertex). The xx-coordinate of the vertex can be found using the formula for the axis of symmetry, because the axis of symmetry intersects the parabola at its vertex. x=-b2a x=\text{-} \frac{b}{2a} Rearranging the function rule so it is shown in standard form gives f(x)=3x0.5x2f(x)=-0.5x2+3x. f(x)=3x-0.5x^2 \quad \leftrightarrow \quad f(x)=\text{-} 0.5x^2+3x. Thus, a=-0.5a=\text{-} 0.5 and b=3.b=3. These values can be substituted into the formula.
x=-b2ax=\text{-}\dfrac{b}{2a}
x=-32(-0.5)x=\text{-}\dfrac{{\color{#009600}{3}}}{2 ({\color{#0000FF}{\text{-} 0.5}})}
x=-3-1x=\text{-}\dfrac{3}{\text{-} 1}
x=31x=\dfrac{3}{1}
x=3x=3
x=3x=3 means that the point on the parabola whose xx-coordinate is 33 is the maximum. In the table, we found that when x=3,y=4.5.x=3, y=4.5. Thus, the height of the tunnel is 4.54.5 meters.
info Show solution Show solution
Method

Graphing a Quadratic Function in Standard Form

It is possible to graph a quadratic function in standard form by using the characteristics given by its function rule. Consider the function f(x)=x24x+3. f(x)=x^2-4x+3.

1

Identify and graph the axis of symmetry
The axis of symmetry can be found using aa and b.b. Here, a=1a=1 and b=-4.b=\text{-}4.
x=-b2ax=\text{-}\dfrac{b}{2a}
x=--421x=\text{-}\dfrac{{\color{#009600}{\text{-}4}}}{2\cdot{\color{#0000FF}{1}}}
x=--42x=\text{-}\dfrac{\text{-}4}{2}
x=-(-2)x=\text{-}(\text{-}2)
x=2x=2
The axis of symmetry for this function is x=2.x=2.


2

Determine and plot the vertex
The axis of symmetry always intersects the parabola at its vertex. Thus, x=2x=2 is the xx-coordinate of the vertex. Use this to determine the corresponding yy-coordinate.
f(x)=x24x+3f(x)=x^2-4x+3
f(2)=2242+3f({\color{#0000FF}{2}})={\color{#0000FF}{2}}^2-4\cdot{\color{#0000FF}{2}}+3
f(2)=48+3f(2)=4-8+3
f(2)=-1f(2)=\text{-}1
The vertex of the function is (2,-1).(2,\text{-}1).

3

Determine and plot the yy-intercept

The yy-intercept can be found using c.c. In this case c=3,c=3, meaning that the yy-intercept is (0,3).(0,3). Add this point to the graph.

4

Reflect the yy-intercept across the axis of symmetry

Since the axis of symmetry divides the graph into two mirror images, there exists another point on the other side of the axis of symmetry with the same yy-value as the yy-intercept. These points are equidistant from the axis of symmetry.

5

Draw the parabola

Now, the general shape of the parabola can be seen. Connect the points with a smooth curve.

Method

Solving Quadratic Equations Graphically

If the dependent variable of a quadratic function is exchanged for a constant, say D,D, the result is a quadratic equation: ax2+bx+c=D. ax^2+bx+c=D.

This type of equation can be solved graphically. This is done by first plotting the function y=ax2+bx+c,y = ax^2+bx+c, then finding the xx-coordinate of the point(s) on the graph that has the yy-coordinate D.D. The xx-coordinate(s) is the solution to the equation.
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