Solution
Step-by-step explanation:
we have:
Part 1) Find the x-intercepts of the parabola and write them as ordered pairs
The x-intercepts are the values of x when the value of y is equal to zero
so
For y=0
[tex](x-4)(x+2)[/tex]For x=4 and x=-2 the equation is equal to zero
therefore
The x-intercepts are the points (-2,0) and (4,0)
Part 2) Write the equation y=(x-4)(x+2) in standard form
The quadratic equation in standard form is equal to
[tex]y=ax^2+bx+c[/tex]applying distributive property
[tex]\begin{gathered} y=(x-4)(x+2) \\ y=x^2+2x-4x-8 \\ y=x^2-2x-8 \end{gathered}[/tex]where a = 1, b = -2, c = -8
Part 3) With the standard form of the equation from part ll, use the quadratic formula to identify the x-value of the vertex
we know that
the x-value of the vertex is -b/2a
we have
where a = 1, b = -2, c = -8
substitute
x = -b/2a = -2/2(1) = 1
therefore
The x-coordinate of the vertex is 1
Part 4) Substitute the x-value of the vertex from part lll into the original equation to find the y-value of the vertex.
we have
[tex]y=(x-4)(x+2)[/tex]For x=1
substitute and solve for y
[tex]\begin{gathered} y=(1-4)(1+2) \\ y=-3(3) \\ y=-9 \end{gathered}[/tex]therefore
The y-coordinate of the vertex is -9
The coordinate of the vertex is the point (1,-9)
