Question

Graph a parabola whose intercepts are x=-3 and x=5 and whose minimum value is y =-4

Answer 1

`y = 4(x^(2) + 8x + 15) = 4x^(2) + 32x + 60`

Explanation:

A parabola has the following format:

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

If a is positive, it's minium value is:

`y_(M) = -(\bigtriangleup)/(4a)`

In which

`\bigtriangleup = b^(2) - 4ac`

Factoring:

`ax^(2) + bx + c = a(x - x_(1))*(x - x_(2))`, in which
`x_(1) and x_(2)` are the intercepts.

In this question:

`x_(1) = -5, x_(2) = -3`

So

`a(x - x_(1))*(x - x_(2)) = a*(x - (-5))*(x - (-3)) = a*(x+5)*(x+3) = a*(x^(2) + 8x + 15)`

Suppose a = 1, we have:

`x^(2) + 8x + 15`

`\bigtriangleup = 8^(2) - 4*1*15 = 4`

The minimum value will be:

`y_(M) = -(4)/(4) = -1`

We want this minimum value to be -4, which is 4 times the current minimum value, so we need to multiply a by 4. Then

`a = 4`

And the parabola is:

`y = 4(x^(2) + 8x + 15) = 4x^(2) + 32x + 60`