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)=?

Question

myininaya · Answer

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 · Answer

so we have from third equation c=2 right?

myininaya · Answer

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

myininaya · Answer

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 · Answer

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

myininaya · Answer

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

myininaya · Answer

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

myininaya · Answer

evaluate that then you are done

anonymous · Answer

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

jim_thompson5910 · Answer

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

anonymous · Answer

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

jim_thompson5910 · Answer

oh oops, thx for pointing that out

jim_thompson5910 · Answer

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