Over which interval is the graph of f(x) = –x2 + 3x + 8 increasing

Question

How to do this?

Answer 1

This is the equation of a downward opening parabola.

f(x) = -x² + 3x + 8

f(x) = -(x² - 3x   ) + 8

f(x) = -(x² - 3x + 1.5² ) 8 + 1.5²     complete the square

f(x ) = -(x - 1.5) + 10.25

 

The vertex is at (1.5, 10.25).

Inceaseing form the left, decreasing to the right.

Answer 2

 
Over which interval is the graph of f(x) = –x2 + 3x + 8 increasing
How to do this?

 

The graph is increasing when the slope is positive and increasing and is decreasing when the slope is negative and increasing

 

find the derivative f'(x)=-2x+3, the slope of the curve is zero at

-2x+3=0, -2x=-3, x=3/2 now x is a critical point

find f(3/2)= –(3/2)² + 3(3/2) + 8 = -9/4+9/2+8= 41/4 therefore at point (3/2,41/4) the slope of the curve is zero this could be either a maximum or a minimum point

 

to determine what is going on before and after the point find

 

f'(1)=-2(1)+3=1>0  the slope is positive for values < 3/2

f"(1)=-2 < 0 Curve is concave down

 

f'(2)=-2(2)+3=-1<0 the slope is negative for values > 3/2

f"(2)=-2 < 0 Curve is concave down 

 

therefore point (3/2,41/4) is a maximum, and the graph of the function is increasing for values < 3/2 where the slope is positive therefore the interval where the function is increasing is (-∞,3/2)