Answer:
A)
(1, 0) and (5, 0)
B)
Yes.
The equation is x = 3.
C)
Yes.
In this case, we have a minimum.
Step-by-step explanation:
We are given the function:
[tex]f(x)=x^2-6x+5[/tex]
Since the highest degree is two, we can immediately determine this to be a quadratic.
A)
The x-intercepts of a function are whenever the function equates to 0. Hence:
[tex]0=x^2-6x+5[/tex]
We can factor:
[tex](x-1)(x-5)=0[/tex]
Zero Product Property:
[tex]x-1=0\text{ or } x-5=0[/tex]
Solving yields:
[tex]x=1 \text{ and } x=5[/tex]
So, our x-intercepts are at (1, 0) and (5, 0).
B)
Since this is a quadratic, it indeed has a line of symmetry.
Recall that the line of symmetry for a quadratic equivalent to the x-coordinate of the vertex.
In the given function, a = 1, b = -6, and c= 5.
Hence, the x-coordinate of the vertex is:
[tex]\displaystyle x=-\frac{b}{2a}=-\frac{(-6)}{2(1)}=\frac{6}{2}=3[/tex]
So, the line of symmetry is x = 3.
C)
Every parabola has a minimum/maximum value.
This depends on the sign of the leading coefficient.
If the leading coefficient is positive, then we have a minimum since our parabola will be curving upwards.
And if the leading coefficient is negative, then we have a maximum since our parabola will be curving downwards.
Since our leading coefficient here is 1, hence positive, we have a minimum value.
