What is the maximum vertical distance between the line
y = x + 72
and the parabola
y = x2 for −8 ≤ x ≤ 9?
what is the maximum vertical distance between the line
y = x + 72
and the parabola
y = x^2 for −8 ≤ x ≤ 9?

the maximum distance is and can be found at x = 

Let's call:

f(x) = x + 72

g(x) = x²

A point belonging to the line will be L(x, x+72) and a point on the parabola will be P(x, x²). It can be easily seen that in the interval -8 ≤ x ≤ 9 the line is above the parabola (it's enough to graph them or plug in some numbers), therefore their distance at any point will be:

d(x) = f(x) - g(x) = - x² + x + 72

The function d(x) is a parabola that opens downward, therefore the maximum will be the vertex; given a parabola

y(x) = ax² + bx + c

the coordinates of the vertex will be

Therefore:

Hence, the maximum distance is  = 72.25 and can be found at x = 

Another way to find the maximum is to use calculus to find the first derivative of the distance:

d'(x) = -2x + 1

and set it equal to zero:

-2x + 1 = 0

Since the second derivative:

d"(x) = -2 

is negative, the point is a maximum.

Then, substitute this value in the equation for the distance:

Hence, the maximum distance is  = 72.25 and can be found at x = 

options a,b,c are the right answer.

step-by-step explanation:

a linear parent function is always represented as y = x

if we write this function in the form of y=mx+c

⇒

or  

now if we plot a graph we will find these characteristics:

1) y intercept of the graph is zero so its a straight line passing through origin.

2) as we can see in both the forms of function written, values of x and y may be either positive or negative for the positive slope means points on the graph are either in first quadrant or in third one.

3) it's slope is one.

c) yes, because of asa or aas