algebra - vertex & intercepts?

y intercept is the easy one set x = 0 and solve

y = 0^2 + 4(0) - 5

y = -5

(0,-5)

for the x ints first try to factor

(x-1)(x+5)

(had factoring not worked you would be forced to use the quadratic equation)

now set the inside of each parenthesis to 0

x-1 = 0

x = 1

x + 5 = 0

x = -5

(1,0) (-5,0)

vertex is -b/2a

-4/2(1)

-2 for the x substitute -2 and solve for y

(-2)^2 + 4(-2) -5

4 - 8 - 5

-9

vertex @ (-2,-9)

To find the y-intercepts, set x = 0, to find (0,-5)

To find the x-intercepts, set y = 0, to get:

x² + 4x - 5 = 0

(x + 5)(x - 1) = 0

so, the y-intercepts are (-5,0), (1,0).

To find the vertex, an easy way is to use the axis of symmetry:

x = -b/(2a) = -4/(2(1)) = -2

f(-2) = -9 and so the vertex is at the point (-2,-9)

f(x) = (x + 2)² - 9

=> the vertex is at (-2, -9)

f(x) = 0 gives the x-intercepts

i.e. (x + 2)² - 9 = 0

=> (x + 2)² = 9

=> x + 2 = ± 3

so, x = -5 or x = 1....i.e. points (-5, 0) and (1, 0)

f(0) gives the y-intercept

=> f(0) = -5....so, point (0, -5)

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