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If an quadratic equation $f(x)$ has max value $4$ when $x = 0$ and $f(-2)=0$, what is the value of $f(3) =$ ?

3 Answers 3

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making the ansatz $$f(x)=ax^2+c$$ we get easily $$c=4$$ from $$f(-2)=0$$ we get $$a=-1$$ and our searched function is $$f(x)=-x^2+4$$

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    $$b$$ must be zero since the extremum can we find for $x=0$ the y-axes is the symmetry line – 2017-01-06
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    I get it I get it I get it get it >>.. I was so close to the answer .... I am so sleepy .----------> Thank you anyway ! – 2017-01-06
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    at first thinking and then computing has told my doctor father – 2017-01-06
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Since the parabola is symmetric about the axis of symmetry, $2$ is another root of the polynomial (draw a picture). Hence the quadratic function can be written as $$ f(x)=a(x+2)(x-2) $$ where $a<0$ (why?). You can solve for $a$ since $f(0)=4$, and then find $f(3)$.

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A quadratic function has the form $a\cdot (x-b)^2 + c$ Given the maximum value at $x=0$ and a negative opening, says $a<0$ and $b=0$.

From there, it should be easy to determine $a$ and $c$ from the two points given. After that, you plug $x=3$ in there.