Answer:
First off, we look for which circles are open or closed.
We start with an open interval since the circle on the left is open and end with a closed interval since the circle on the right is closed.
Domain is all x values, Range is all y values
The graph shows the continous function going from -3 to 1 on the x axis.
According to the circles, this means our domain will be (-3,1].
Now, the range doesn't care about if its closed or not. So we can say the graph is on the y axis from -4 and 0. This means the range is -4<y<0
I used different notations for both just incase you need to represent your answer differently :)
-3<x<1 & (-3,1] . Range is [-4,0]. 0>y>-4 looks correct as-well.
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.

The 2 and one-half cancel each other out.


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