SOLUTION: How do I find the zeroes of a function of f(x)= -x squared plus 3x plus 6 ?

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Question 187133This question is from textbook Precalculus
: How do I find the zeroes of a function of f(x)= -x squared plus 3x plus 6 ? This question is from textbook Precalculus

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=+-x%5E2%2B3x%2B6 Start with the given function


0=-x%5E2%2B3x%2B6 Plug in f%28x%29=0


0=x%5E2-3x-6 Multiply EVERY term by -1 to make the leading coefficient positive.


Notice we have a quadratic equation in the form of ax%5E2%2Bbx%2Bc where a=1, b=-3, and c=-6


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%28-3%29+%2B-+sqrt%28+%28-3%29%5E2-4%281%29%28-6%29+%29%29%2F%282%281%29%29 Plug in a=1, b=-3, and c=-6


x+=+%283+%2B-+sqrt%28+%28-3%29%5E2-4%281%29%28-6%29+%29%29%2F%282%281%29%29 Negate -3 to get 3.


x+=+%283+%2B-+sqrt%28+9-4%281%29%28-6%29+%29%29%2F%282%281%29%29 Square -3 to get 9.


x+=+%283+%2B-+sqrt%28+9--24+%29%29%2F%282%281%29%29 Multiply 4%281%29%28-6%29 to get -24


x+=+%283+%2B-+sqrt%28+9%2B24+%29%29%2F%282%281%29%29 Rewrite sqrt%289--24%29 as sqrt%289%2B24%29


x+=+%283+%2B-+sqrt%28+33+%29%29%2F%282%281%29%29 Add 9 to 24 to get 33


x+=+%283+%2B-+sqrt%28+33+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%283%2Bsqrt%2833%29%29%2F%282%29 or x+=+%283-sqrt%2833%29%29%2F%282%29 Break up the expression.


So the answers are x+=+%283%2Bsqrt%2833%29%29%2F%282%29 or x+=+%283-sqrt%2833%29%29%2F%282%29


which approximate to x=4.372 or x=-1.372