SOLUTION: Really stuck on this one! For the quadratic function y =F(x) = x^2-2x-15 i need the vertex and x & Y intercepts, graph, domain & range..thank you so much, Carol

Really stuck on this one! For the quadratic 
function y  = f(x) = x^2-2x-15 i need the 
vertex and x & Y intercepts, graph, domain 
& range..thank you so much,  Carol


The x-coordinate of the vertex of the parabola 

          y = f(x) = ax² + bx + c 

is h = -b/(2a)

And the y-coordinate of the vertex is

k = f(h)

Its domain is (-¥, ¥)

and the range is 

[k, ¥)  if a > 0

and 

(-i, k]  if a < 0

The y-intercept is (0, c)

The x intercepts are the (r1,0), (r2,0)

where r1 and r2 are the solutions to the

equation ax² + bx + c = 0 if they are 

real numbers.  If they are imaginary

numbers there are no x intercepts.


Your problem is y = f(x) = x² - 2x - 15

so a = 1, b = -2, c = -15 

The x-coordinate of the vertex of the parabola 

          y = f(x) = x² - 2x - 15 

is h = -b/(2a) = -(-2)/[2(1)] = 1

And the y-coordinate of the vertex is

k = f(h) = f(1) = (1)² - 2(1) - 15 = -16

So the vertex is (h, k) = (-1, -16)

Its domain is (-¥, ¥)

and the range is 

[k, ¥)  if a > 0

and since a = 1, which is > 0 and k = -16 the 
range is [-16, ¥	)

The y-intercept is (0, c)

and since c = -15, the y-intercept is (0, -15)

The x intercepts are (r1,0), (r2,0)

where r1 and r2 are the solutions to the

equation ax² + bx + c = 0 if they are 

real numbers.  

Solve this equation: 

 x² - 2x - 15 = 0

you will get solutions -3 and 5 

so the x-intercepts are (-3,0) and (5,0)

The graph is:

  

Edwin