Question

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

Answer 1

Answer:

$y = 4(x^{2} + 8x + 15) = 4x^{2} + 32x + 60$

Step-by-step explanation:

A parabola has the following format:

$y = f(x) = ax^{2} + bx + c$

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

$y_{M} = -\frac{\bigtriangleup}{4a}$

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:

$y_{M} = -\frac{4}{4} = -1$

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:

$y = 4(x^{2} + 8x + 15) = 4x^{2} + 32x + 60$