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

2 Answers
Apr 2, 2017

f(x) = -x^2-6xf(x)=x26x

Explanation:

f(x) = ax^2+bx+cf(x)=ax2+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)

Apr 2, 2017

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)

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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

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