Answer:
They have the same x-value
f(x) has the greater minimum
Step-by-step explanation:
To find the vertex of a second degree equation, in this case the minimum value, we can use the following equation:
x = -b / 2a
Remember that a second degree equation has the following form:
ax^2 + bx + c
so a = 1, b = -8 and c = 7. Now you have to substitute in the previous equation
x = - (-8) / 2(1)
x = 8 / 2
x = 4
This means that the two functions have the same x-value.
The y value of f(x) would be
f(4) = (4)^2 - 8(4) + 7
f(4) = 16 - 32 + 7
f(4) = -9
So the vertex, or minimun value of f(x) would be at the point (4, -9).
The vertex, or minimun value of g(x) is at the point (4, -4).
So f(x) has a minimum value of -9 and g(x) a minimum value of -4.
Answer:
[tex]f(x)[/tex] has the greater minimum at [tex](4,-9)[/tex]
Step-by-step explanation:
The given function is
[tex]f(x)=x^{2}-8x+7[/tex]
Additionally, the graph attached shows the function [tex]g(x)[/tex] which as a minimum at (4,-4).
So, let's find the minimum of [tex]f(x)[/tex], which has coordinates [tex](h,k)[/tex] where [tex]h[/tex] is defined as
[tex]h=-\frac{b}{2a}[/tex]
Where [tex]a=1[/tex] and [tex]b=-8[/tex], replacing these values, we have
[tex]h=-\frac{b}{2a}=-\frac{-8}{2(1)}=4[/tex]
So, the other coordinate is defined as [tex]k=f(h)[/tex], so replacing the value in the function, we have
[tex]f(x)=x^{2}-8x+7\\f(4)=(4)^{2}-8(4)+7=16-32+7=-9[/tex]
So, the minimum of [tex]f(x)[/tex] is at [tex](4,-9)[/tex].
Therefore, [tex]f(x)[/tex] has the greater minimum at [tex](4,-9)[/tex]
