Find the x and y-intercepts of the following equation: y = x^2 + 3x - 40?

Question

Really urgent! thanks all =)

Answer 1

The y-intercept occurs where x = 0, so:

y = x^2 + 3x - 40
y = 0^2 + 3*0 - 40
y = 0 + 0 - 40
y = -40

Y-intercept at (0,-40)


The x-intercept occurs where y = 0, so:

y = x^2 + 3x - 40
0 = x^2 + 3x - 40
0 = (x+8)(x-5)



X-intercepts at (-8,0) and (5,0)

Answer 2

positive 2nd degree, so it is a parabola concave up which has 2 x-values.

for y-intercept, set x=0:
y=0+0-40
y=-40
(0,-40)

for x-intercept, set y=0:
0=x^2+3x-40
0=(x+8)(x-5)
x+8=0 or x-5=0
x=-8 or x=5
(-8,0) or (5,0)

Answer 3

y = x^2 + 3x - 40

y = (x+8)(x-5) x intercepts at x= -8 y=0 and x =5 y=0

minimum at (-1.5 , -44.275)

Answer 4

y = x^2 + 3x - 40
y = (x + 8) (x - 5)

Answer: (- 8, 0) (5, 0)