Question

Consider the two functions.

Answer 1

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 2

Answer:

$f(x)$ has the greater minimum at $(4,-9)$

Step-by-step explanation:

The given function is

$f(x)=x^{2}-8x+7$

Additionally, the graph attached shows the function $g(x)$ which as a minimum at (4,-4).

So, let's find the minimum of $f(x)$, which has coordinates $(h,k)$ where $h$ is defined as

$h=-\frac{b}{2a}$

Where $a=1$ and $b=-8$, replacing these values, we have

$h=-\frac{b}{2a}=-\frac{-8}{2(1)}=4$

So, the other coordinate is defined as $k=f(h)$, so replacing the value in the function, we have

$f(x)=x^{2}-8x+7\\f(4)=(4)^{2}-8(4)+7=16-32+7=-9$

So, the minimum of $f(x)$ is at $(4,-9)$.

Therefore, $f(x)$ has the greater minimum at $(4,-9)$