Category Archives: Sullivan Chapter 4

Application of Quadratic Function: Maximize

Example:

A baseball player swings and hits a pop fly straight up in the air to the catcher.  The height of the baseball in meters t seconds after it is hit is given by the quadratic function h(t)=-4.9t^2+17.4t+1.  How long does it take the baseball to reach its maximum height?  What is the maximum height obtained by the baseball?

Solution:

h(t)=-4.9t^2+14.7t+1 is defined to be a quadratic function.  The graph of a quadratic function is a parabola.  This quadratic function has a leading coefficient of -4.9 which means that the parabola is opening down.

Find the vertex to find the maximum value.

The formula for the x coordinate of the vertex
t=-b/{2a}
For a quadratic function a is the coefficient of the square term, b is the coefficient of the linear term, and c is the constant
h(t)=-4.9t^2+14.7t+1
a=-4.9; b=14.7; c=1
Substitute the values of a and b into the formula
t=-b/{2a}
t={-(14.7)}/{2(-4.9)}
Simplify with a calculator
t={-(14.7)}/{2(-4.9)}
t=1.5

The ball reaches the maximum height 1.5 seconds after the ball was hit.

The maximum height can be found by substituting 1.5 seconds for time in the height function.

The height function
h(t)=-4.9t^2+14.7t+1
Substitute t=1.5 in the height function
h(t)=-4.9t^2+14.7t+1
h(1.5)=-4.9(1.5)^2+14.7(1.5)+1
Simplify using a calculator
h(1.5)=-4.9(1.5)^2+14.7(1.5)+1
h(1.5)=12.025

The maximum height of the ball is 12.025 meters.

 

Quadratic Function

Example: Rewrite the given quadratic function in standard form by completing the square.  Then state the domain, range, vertex, x-intercepts, y-intercept, the orientation (opens up or opens down), and the axis of symmetry.  Finally graph the function.

f(x)=-x^2-2x+8

Solution:

Rewrite function in standard form.

f(x)=-x^2-2x+8

f(x)=-(x^2+2x)+8

f(x)=-(x^2+2x+1)+8+1

f(x)=-(x+1)^2+9

Here is a youTube video that demonstrations the process.

Find the vertex.

When the quadratic function is written in standard form you can identify the vertex as (h,k).

f(x)=a(x-h)^2+k

f(x)=-(x+1)^2+9

The vertex of the quadratic function is (-1,9).

Find the orientation.

The leading coefficient of the quadratic function is negative so the parabola opens down.

Find the y-intercept.

To find a y-intercept let x=0.

f(0)=-(0+1)^2+9

f(0)=-(1)^2+9

f(0)=-1+9

f(0)=8

The y-intercept is (0,8).

Find the x-intercept.

To find an x-intercept let y=0 or f(x)=0.

0=-(x+1)^2+9

0-9=-(x+1)^2+9-9

-9=-(x+1)^2

{-9}/{-1}={-(x+1)^2}/{-1}

9=(x+1)^2

sqrt{9}=sqrt{(x+1)^2}

pm 3=x+1

-1 pm 3=x+1-1

-1 pm 3=x

-1 + 3=x or -1 - 3=x

2=x or -4=x

Find the axis of symmetry.

The axis of symmetry is a vertical line that passes through the vertex.  Since the vertex is (-1,9) the axis of symmetry is x=-1.

Graph the function.

Plot the intercepts and the vertex.

Since the vertex is the highest point we can draw the parabola using the peak at the vertex.

Find the domain and range.

The domain is (- infty, infty).

The range is (  - infty, 9 {}{]}.