we see the quadratic coefient is positive so it opens up
so the vertex is the minimum
so
hack
for f(x)=ax²+bx+c
the x value of the vertex is -b/2a
also that is the axis of symmetry
given
h(x)=1x²-6x-7
x value of vertex is -(-6)/(2*1)=6/2=3
that isi also axis of symmetr, x=3
to find y value evaluage h(3)
h(3)=3²-6(3)-7
h(3)=9-18-7
h(3)=-16
vertex is (3,-16)
it is a minimum
the axis of symmetry is x=3
x intercept is where the line crosses the x axis or where y=0 or h(x)=0
h(x)=0 solve
0=x²-6x-7
0=(x-7)(x+1)
x-7=0
x=7
x+1=0
x=-1
x intercepts at x=-1 and x=7 or the points (-1,0) and (7,0)
y intercept is where line crosses y axis or where x=0
evaluate h(0)
h(0)=0²-6(0)-7
h(0)=-7
y int at y=-7 or point (0,-7)
A. 1
B. 3
C. 4
D. 5
E. 2