Question 91184
First equation

a) The equation is already in the form you need.
b)  There is NO vertex in its graph since it is a linear equation , the graph is a straight line.

Second equation

{{{y = x^2-2x-8}}}

a)
For y intercept x = 0, so let x =0 in the equation to calculate y value

{{{y = 0^2 - 2*0 - 8}}}

= -8
so y intercept point is (0, -8).

b)Similarly, let y = 0, we have {{{0 = x^2-2x-8}}}

(x-4)(x+2) = 0

x = 4 or  x = -2

so there are two x intercept points: (4, 0) and (-2, 0).


c)when {{{x = -b/(2a)}}}, the y value will be the smallest, that's vertex

here a = 1, b = -2

so {{{x = -(-2)/2}}}, which is x = 1,

when x = 1, y = -9

so vertex is (1, -9)
 

d. Graph the equation.

{{{graph(200,200, -5,5, -10, 8, x^2-2x-8)}}}