Answer:
y-intercept = (0, 8)
x-intercepts = (4, 0) and (-2, 0)
vertex = (1, 9)
5th point = (2, 8)
Step-by-step explanation:
As the coefficient of x² is negative, the curve will be "n" shaped.
y-intercept
crosses the y-axis when x = 0
substitute x = 0 into the function and solve for y:
[tex]\implies y=-(0)^2+(2 \times 0)+8=8[/tex]
Therefore, y-intercept = (0, 8)
x-intercept
crosses the x-axis when y = 0
set the function to zero and solve for x:
[tex]\implies -x^2+2x+8=0[/tex]
[tex]\implies x^2-2x-8=0[/tex]
[tex]\implies (x-4)(x+2)=0[/tex]
[tex]\implies x=4, x=-2[/tex]
Therefore, x-intercepts = (4, 0) and (-2, 0)
vertex
vertex form: [tex]y=a(x-h)^2 +k[/tex], where (h, k) is the vertex
expand vertex equation: [tex]y=ax^2 -2ahx+ah^2 +k[/tex]
compare coefficients with original function: [tex]y=-x^2+2x+8[/tex]
[tex]\implies a=-1, h=1, k=9[/tex]
So vertex = (1, 9)
5th plot point
You can choose any value of x and input it into the equation for y, but for symmetry, I have chosen to find the other value of x (aside x = 0) when y = 8
[tex]\implies -x^2+2x+8=8[/tex]
[tex]\implies -x^2+2x=0[/tex]
[tex]\implies x^2-2x=0[/tex]
[tex]\implies x(x-2)=0[/tex]
[tex]\implies x=0, x=2[/tex]
5th point = (2, 8)
