Determining Maximum and Minimum Values of a quadratic Function!!

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# Determining Maximum and Minimum Values of a quadratic Function!! - PowerPoint PPT Presentation

Determining Maximum and Minimum Values of a quadratic Function!!. Sometimes there is a little confusion…. When a question asks for the maximum or the minimum of a quadratic function, it is not asking for the whole vertex. It is simply asking for the y coordinate of the vertex

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### Determining Maximum and Minimum Values of a quadratic Function!!

Sometimes there is a little confusion…..
• When a question asks for the maximum or the minimum of a quadratic function, it is not asking for the whole vertex.
• It is simply asking for the y coordinate of the vertex
• It’s important to keep this in mind because when answering word problems etc…..we don’t want to do more work than is necessary.
What equations tell us.
• By looking at an equation in standard form, factored form or vertex form, we can immediately tell whether the parabola will have a max or min.
• Simply look at the “a” value.
• If a > 0 the parabola opens up and we have a minimum value
• If a < 0 the parabola opens down and we have a maximum value.
Finding the max or min value….there are lots of ways.
• If the equation is in standard form we can:
• Complete the square to put it in vertex form
• Express the equation in factored form, find the midpoint of the zeros to get the axis of symmetry etc
• Use quad formula to find the zeros
• Use a graphing calculator (we won’t discuss this here
Let’s start out with completing the square.
• Unfortunately…..there are some things in life that you just have to remember…..
• Procedures, procedures, procedures
• Have you ever watched a pilot before a flight?
• What does he/she do?
• Why?
• After lots of practice does he/she have it pretty well memorized…..you don’t have to think about it anymore hopefully…
Procedures….
• Completing the square algebraically is much like this.
• You have to follow a set of rules to get you to your desired destination.
• Follow along and fill out your spread sheet as we complete the square for the equation y=x2+12x+40
Step 1

Insert brackets around the first two terms

Step 2

Factor out any value in front of the x2

In this case there is nothing to factor out

Step 3

Take Middle term divide by 2 & square it

Step 4

Rewrite the equation with the result from step 3 added and subtracted inside the brackets

Step 5

Bring the (-) term outside of the bracket remember to remultiply if necessary

Step 6

Combine the two constant terms outside the bracket

Step 7

Factor trinomial inside the bracket

Step 8

And that’s all folks……

There isn't much more to this procedure.

Of course there are more difficult equations

But the process essentially remains the same

Recall….
• The factored form of a quadratic equation is
When an equation is in factored form….
• The axis of symmetry is the vertical line that runs through the midpoint of the x intercepts.
• This can be found by the following calculation
• This value is also the x coordinate of the vertex.
When an equation is in factored form
• Recall, you find the y coordinate of the vertex by plugging the x coordinate of vertex into original equation and solving for y.
• So Remember……..
• The x value of the vertex gives the axis of symmetry,
• The ‘y’ value gives the max/min value of the function
EXAMPLE

Find the zeroes, vertex, AOS, max/min for the function

y=3(x-8)(x+5)

• X coordinate of vertex
• Y coordinate of vertex
• Unfortunately we can’t always solve quadratic equations by factoring.
• As a result , a formula has been developed that will always let us solve for the x intercepts
• If f(x) = ax2 + bx + c is given then we could use the quadratic formula to find the roots of the equation.
• The “x” represents the x intercept.
• The “+/-” represents the fact that there could be two x intercepts
• a is the coefficient in front of the x2 term
• b is the coefficient in front of the x term
• c is the number/ constant term at the end
Example solve for the vertex.
• Using the Quadratic Formula solve 2x2 - 5x – 1 = 0

a = 2

b = -5

c= -1

Therefore, there are two solutions

x = 2.69 and x = -0.19

Next….
• Plug this into original equation.
Once more thing
• If your equation is in standard form there is a short cut to getting the x coordinate of the vertex.
For example:
• Find the minimum of
• Now plug this into the original equ:

The minimum is –0.25 and it occurs where x = -3.5