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)=?

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

so we have from third equation c=2 right?

so we have equations \[9a-3b+2=0, 9a+3b+2=0\]

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

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

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

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

evaluate that then you are done

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

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) \[f(x)=\frac{2}{9}x^2-2\] \[f(-1)=\frac{2}{9}(-1)^2-2\] \[f(-1)=\frac{2}{9}(1)-2\] \[f(-1)=\frac{2}{9}-2\] \[f(-1)=-\frac{16}{9}\]

@jim we both made typos it is \[f(0)=2\]

oh oops, thx for pointing that out

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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