Respuesta :
Answer:
[tex]y = 4(x^{2} + 8x + 15) = 4x^{2} + 32x + 60[/tex]
Step-by-step explanation:
A parabola has the following format:
[tex]y = f(x) = ax^{2} + bx + c[/tex]
If a is positive, it's minium value is:
[tex]y_{M} = -\frac{\bigtriangleup}{4a}[/tex]
In which
[tex]\bigtriangleup = b^{2} - 4ac[/tex]
Factoring:
[tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], in which [tex]x_{1} and x_{2}[/tex] are the intercepts.
In this question:
[tex]x_{1} = -5, x_{2} = -3[/tex]
So
[tex]a(x - x_{1})*(x - x_{2}) = a*(x - (-5))*(x - (-3)) = a*(x+5)*(x+3) = a*(x^{2} + 8x + 15)[/tex]
Suppose a = 1, we have:
[tex]x^{2} + 8x + 15[/tex]
[tex]\bigtriangleup = 8^{2} - 4*1*15 = 4[/tex]
The minimum value will be:
[tex]y_{M} = -\frac{4}{4} = -1[/tex]
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
[tex]a = 4[/tex]
And the parabola is:
[tex]y = 4(x^{2} + 8x + 15) = 4x^{2} + 32x + 60[/tex]
