Determining Maximum and Minimum Values of a quadratic Function!!

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.

How is this, you ask? • 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…

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 Express answer in vertex form

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

What about if we want to solve for a max or min without completing the square?

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

What if you can’t factor and you forget how to complete the square?

The Quadratic Formula • 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

The Quadratic Formula • If f(x) = ax2 + bx + c is given then we could use the quadratic formula to find the roots of the equation. • The quadratic formula: • 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

Determining Maximum and Minimum Values of a quadratic Function!!

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