Question

How do you find a quadratic function f(x) = ax² + bx + c given maximum value 9; zeros of f are -6 and 0?

Answer 1

`f(x) = -x^2-6x`

Explanation:

`f(x) = ax^2+bx+c`

`f'(x) = 2ax+b`

`f''(x) = 2a`

We are told that `f(x)` has a maximum value.
`:. f''(x) <0 -> a<0`

We are also told that `f(x)` has zeros of `-6` and `0`

Hence: `(x+6)` is a factor of `f(x)`
Also: `-x` is a factor of `f(x)` (since `a<0)`

`:. f(x) = -x(x+6) -> -x^2-6x`

To check our result, we are told that `f_max = 9`

With our result `f'(x) = -2x-6`

For a critical value `f'(x) = 0 -> -2x-6 =0`
`:. -2x=6 -> x=-3` should give the maximum value of `f(x)`

`f(-3) = 3(-3+6) = 9` which is the given maximum value of `f(x)`

Answer 2

`y=-x^2-6x`

Explanation:

If the vertex is a maximum then the graph is of general shape `nn`
Consequently the coefficient of `x^2` ie `a`, is negative so we have:

`color(brown)(y=-ax^2+bx+c" "->" "y=(-1)ax^2+bx+c)`

`color(brown)("The x-intercepts are at "x=-6 and x=0)`

`x_("vertex")` is half way between these so we have:

`x_("vertex")=(-6)/2=-3`

It is given that `y_("vertex")=9`

`color(brown)("Vertex"->(x,y)=(-3,9)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("We now have 3 points and 3 unknowns so solvable")`

Form the vertex:
`9=-a(-3)^2+b(-3)+c" "->" "9=-9a-3b+c`

From x-intercept `=-6`
`0=-a(-6)^2+b(-6)+c" "->" "0=-36a-6b+c`

From x-intercept `=0`
`0=-a(0)^2+b(0)+c" "->" "color(green)(ul(bar(|color(white)(2/2)c = 0color(white)(2/2)|))`

..........................................................................
Using `c=0` we have:

`9=-9a-3b" "..............Equation(1)`
`0=-36a-6b" ".............Equation(2)`

Consider `Equation(2)`

Write as `+36a=-6b`
divide both sides by `-6`

`36/(-6)a=b`

`b=-6a" "........................Equation(3)`

Using equation(3) substitute for `b` in equation(1)

`9=-9a-3(-6a)`

`9=-9a+18a`

`9=+9a`

`color(green)(ul(bar(|color(white)(2/2)a=1color(white)(2/2)|)`

Using `a=1` substitute for `a` in equation(1)

`9=-9(1)-3b`

Multiply both sides by ( -1 )

`-9=+9+3b`

`-18=3b`

`color(green)(ul(bar(|color(white)(2/2)b=(-18)/3 = -6color(white)(2/2)|)))`
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`y=-ax^2+bx+c" "->" "y=-x^2-6x`