Verified Answer :
AThe function in vertex form if f(x)=(x-4)^2-1
D. The y-intercept of the function is (0,5).
E. The function crosses the x-axis twice.
D. The y-intercept of the function is (0,5).
E. The function crosses the x-axis twice.
Answer:
The vertex of the function is the point \((-3,-13)\)
The graph increase over the interval--------> (-3,∞)
Step-by-step explanation:
we have
\(f(x)=6x-4+x^{2}\)
1) Convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
\(f(x)+4=x^{2}+6x\)
Complete the square. Remember to balance the equation by adding the same constants to each side
\(f(x)+4+9=x^{2}+6x+9\)
\(f(x)+13=x^{2}+6x+9\)
Rewrite as perfect squares
\(f(x)+13=(x+3)^{2}\)
\(f(x)=(x+3)^{2}-13\) -----> function in vertex form
2) Find the vertex
The vertex of the function is the point \((-3,-13)\)
3) Find the axis of symmetry
we know that
In a vertical parabola, the axis of symmetry is equal to the x-coordinate of the vertex
The x-coordinate of the vertex in this problem is equal to \(x=-3\)
therefore
the equation of the axis of symmetry is \(x=-3\)
4) Find the increase-decrease intervals
The graph increase over the interval--------> (-3,∞)
The graph decrease over the interval--------> (-∞,-3)
see the attached figure to better understand the problem
5) Find the x-intercepts of the function
we know that
the x-intercepts are the values of x when the value of the function is equal to zero
In this problem the x-intercepts are
\((-6.61,0)\) and \((0.61,0)\)
so
The function cross the x-axis twice
see the attached figure