SOLUTION: graph f(x)= x(x-2)(x+3) d = -5 < x <4 Find: 1. degree 2. zeros 3. x - intercept 4. y - intercept 5. turning point

Algebra ->  Graphs -> SOLUTION: graph f(x)= x(x-2)(x+3) d = -5 < x <4 Find: 1. degree 2. zeros 3. x - intercept 4. y - intercept 5. turning point      Log On


   



Question 981122: graph f(x)= x(x-2)(x+3)
d = -5 < x <4
Find:
1. degree
2. zeros
3. x - intercept
4. y - intercept
5. turning point

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Your "d" is not really involved in the function. If "d" is for domain, know that your function is not restricted to that set of x values.

f%28x%29=+x%28x-2%29%28x%2B3%29
Polynomial function of degree 3, so since it is also in factored form, you see the roots immediately.

x-intercepts -3, 0, 2.

Like an S shape, maybe a maximum hump and a minimum hump. You did not say if you want to identify the turning points nor if you know about derivatives.

y-intercept: occurs at x=0, so this will be (0,0).

f%3C0 for x%3C-3;
f%3E0 for -3%3Cx%3C0;
f%3C0 for 0%3Cx%3C2;
f%3E0 for x%3E2.

The graph is failing to appear...
graph(300,300,-5,5,-8,8,x(x+3)(x-2))
graph%28300%2C300%2C-4%2C4%2C-6%2C10%2Cx%5E3%2Bx%5E2-6x%29