Respuesta :
The vertex form is a(x-h)²+k, where (h,k) are the coordinates of the vertex.
[tex]f(x)= \\ 5x^2+10x+8= \\ 5(x^2+2x)+8= \\ 5((x^2+2x+1)-1)+8= \\ 5((x+1)^2-1)+8= \\ 5(x+1)^2-5+8= \\ 5(x+1)^2+3[/tex]
The coordinates of the vertex are (-1,3).
The coefficient of x² is positive, so the parabola opens upwards and the vertex is the minimum of the function.
The answer is d.
[tex]f(x)= \\ 5x^2+10x+8= \\ 5(x^2+2x)+8= \\ 5((x^2+2x+1)-1)+8= \\ 5((x+1)^2-1)+8= \\ 5(x+1)^2-5+8= \\ 5(x+1)^2+3[/tex]
The coordinates of the vertex are (-1,3).
The coefficient of x² is positive, so the parabola opens upwards and the vertex is the minimum of the function.
The answer is d.
Answer : d. minimum (-1,3)
[tex]f(x) = 5x^2 + 10x + 8[/tex]
The vertex form of quadratic function is
[tex]f(x) = a(x-h)^2 + k[/tex], where (h,k) is the vertex
To get vertex form we apply completing the square method
To apply completing the square method , there should be only x^2
So we factor out 5 from from first two terms
[tex]f(x) = 5(x^2 + 2x) + 8[/tex]
Now we take the number before x (coefficient of x) and divide by 2
[tex]\frac{2}{2}[/tex] =1
Now square it
[tex](1)^2 =1[/tex]
Add and subtract 1 inside the parenthesis
[tex]f(x) = 5(x^2 + 2x + 1 - 1) + 8[/tex]
Now we take out -1 by multiplying 5
[tex]f(x) = 5(x^2 + 2x + 1) -5 + 8[/tex]
[tex]f(x) = 5(x^2 + 2x + 1) + 3[/tex]
Now we factor x^2 +2x+1 as (x+1)(x+1)
[tex]f(x) = 5(x+1)(x+1) + 3[/tex]
[tex]f(x) = 5(x+1)^2 + 3[/tex]
h=-1 and k=3
So vertex is (-1,3)
When the value of 'a' is negative , then it is a maximum
When the value of 'a' is positive , then it is a minimum
[tex]f(x) = 5x^2 + 10x + 8[/tex] is in the form of [tex]f(x) = ax^2 + bx + c[/tex]
The value of a is 5
5 is positive so it is a minimum
f(x) is minimum at point (-1,3)
