Question

How to solve these question?

Answer 1

3.) An extreme value refers to a point on the graph that is possibly a maximum or minimum. At these points, the instantaneous rate of change (slope) of the  graph is 0 because the line tangent to the point is horizontal. We can find the rate of change  by taking the derivative of the function.

y' = 2ax + b

Now that we where the derivative, we can set it equal to 0.

2ax + b = 0

We also know that at the extreme value, x = -1/2. We can plug that in as well.

$2a (-\frac{1}{2} ) + b = 0$

The 2 and one-half cancel each other out.

$-a + b = 0$

$a = b$

Now we know that a and b are the same number, and that ax^2 + bx + 10 = 0 at x = -1/2. So let's plug -1/2 in for x in the original function, and solve for a/b.

a(-0.5)^2 + a(-0.5) + 10 = 0

0.25a - 0.5a + 10 = 0

-0.25a = -10

a = 40

b = 40

To determine if the extrema is a minima or maxima, we need to go back to the derivative and plug in a/b.

80x + 40

Our critical number is x = -1/2. We need to plug a number that is less than -1/2  and a number that is greater than -1/2 into the derivative.

LESS THAN:
80(-1) + 40 =  -40

GREATER THAN:
80(0) + 40 = 40

The rate of change of the graph changes from negative  to positive at x = -1/2, therefore the extreme value is a minimum.

4.) If the quadratic function is symmetrical about x = 3, that means that the minimum or maximum must be at x = 3.

y' = 2ax + 1

2a(3) + 1 = 0

6a = -1

a = -1/6

So now plug the a value and x=3 into the original function to find the extreme value.

(-1/6)(3)^2 + 3 + 3 = 4.5

The extreme value is 4.5