please help with #28 and #32!

Question

For the exercise, evaluate f at the indicated values f(-3), f(2), f(-a), -f(a), f(a + h).

28. f(x) =-5x2 + 2x-1

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32. Given the function g(x) = 5 - x^2, evaluate g(x+h)-g(x)/h. h ‡ 0.

Answer 1

Problem 28:

For f(-3), you plug in x = -3 and do the arithmetic: f(-3) = -5(-3)² + 2(-3) - 1 = -45 - 6 - 1 = -52

For f(-a) you would plug in x = -a: f(-a) = -5(-a)² + 2(-a) - 1 = -5a² - 2a - 1

For f(a + h), again, plug in x = (a + h):

f(a+h) = -5(a+h)² + 2(a+h) - 1 = -5(a² + 2ah + h²) + 2a + 2h - 1 = -5a² - 10ah - 5h² + 2a + 2h - 1

Problem 32:

g(x) = 5 - x²

g(x+h) = 5 - (x + h)² = 5 - (x² + 2xh + h²) = 5 - x² - 2xh - h²

So g(x+h) - g(x) = (5 - x² - 2xh - h²) - (5 - x²) = 5 - x² - 2xh - h² - 5 + x² = -2xh - h² = h(-2x - h)

The dividing by h gives: h(-2x - h)/h = -2x - h

Answer 2

For the exercise, evaluate f at the indicated values f(-3), f(2), f(-a), -f(a), f(a + h).

28. f(x) =-5x2 + 2x-1

You should simply replace x with the number in each f(). I'm assuming -5x2 is -5x²

f(-3) = -5(-3)²+2(-3)-1 = -5*9-6-1 = -35-7 = -42

f(2) = -5(2)² +2(2) -1 = -5*4+4-1 = -20+3 = -17

f(-a) = -5(-a)² + 2a - 1 = -5a²+2a-1

-f(a) = -[-5a² +2a -1] = 5a² -2a +1

f(a+h) = -5(a+h)² + 2(a+h) - 1 = -5(a² + 2ah+ h²) +2a + 2h - 1 = -5a² -10ah -5h² + 2a + 2h -1

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32. Given the function g(x) = 5 - x^2, evaluate g(x+h)-g(x)/h. h ‡ 0.

I'm assuming it should be [g(x+h) - g(x)]/h

Same concept here: replace x with x+h

g(x+h) = 5-(x+h)² = 5 - (x² + 2xh + h²) = 5- x² -2xh - h²

g(x) = (5-x²)

[g(x+h)-g(x)]/h = [5- x² -2xh - h² - (5 -x²)]/h =

= [5- x² -2xh - h² - 5+ x²]/h

x² and 5 terms cancel, left with:

= [-2xh-h²]/h

= -2x-h

Hope this helps!