Mathematics
OpenStudy (anonymous):

If the function f(x)=ax^2+bx+c has x-intercepts (-3,0) and (3,0) and y-intercept (0,2), then f(-1)=?

myininaya (myininaya):

f(-3)=a(-3)^2+b(-3)+c=0 f(3)=a(3)^2+b(3)+c=0 f(0)=a(0)^2+b(0)+c=2 is given

myininaya (myininaya):

so we have from third equation c=2 right?

myininaya (myininaya):

so we have equations $9a-3b+2=0, 9a+3b+2=0$

myininaya (myininaya):

so we have two equations and two unknowns so we can do 9a-3b+2=0 -(9a+3b+2=0) -------------- 0 -6b+0=0 -6b=0 b=0

myininaya (myininaya):

9a-3b+2=0 we have 9a+2=0 9a=-2 a=-2/9

myininaya (myininaya):

so $f(x)=\frac{-2}{9}x^2+2$

myininaya (myininaya):

$f(-1)=\frac{-2}{9}(-1)^2+2$

myininaya (myininaya):

evaluate that then you are done

OpenStudy (anonymous):

$f(x)=a(x-3)(x+3)$ $f(0)=-9a=0$ $a=-\frac{2}{9}$ $f(x)=-\frac{2}{9}(x+3)(x-3)$

jimthompson5910 (jim_thompson5910):

Alternative route: x=3 or x=-3 x-3=0 or x+3=0 a(x-3)(x+3)=0 a(x^2-9)=0 ax^2-9a=0 Since f(0)=-2, this means that when x=0, then a(0)^2-9a=-2. Solve for a to get a = 2/9 So the function is $f(x)=\frac{2}{9}x^2-9(\frac{2}{9})$ which simplifies to $f(x)=\frac{2}{9}x^2-2$ Finally, plug in x=-1 to evaluate f(-1) $f(x)=\frac{2}{9}x^2-2$ $f(-1)=\frac{2}{9}(-1)^2-2$ $f(-1)=\frac{2}{9}(1)-2$ $f(-1)=\frac{2}{9}-2$ $f(-1)=-\frac{16}{9}$

OpenStudy (anonymous):

@jim we both made typos it is $f(0)=2$

jimthompson5910 (jim_thompson5910):

oh oops, thx for pointing that out

jimthompson5910 (jim_thompson5910):

so that means that the function is $\large f(x)=-\frac{2}{9}x^2+2$ and $\large f(-1)=\frac{16}{9}$

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