Answer:
y=1/2-2
Step-by-step explanation:
Answer:
Domain= (-inf,inf)/(-∞,∞)
Range= (-inf,25/16]
x-intercepts= (0,0), (5/4,0)
y-intercepts=(0,0)
vertex= maximum (5/8,25/16)
Answer:
I'm sorry I don't know if I do get an idea I'll come back and edit
Step-by-step explanation:
I’d need to know the question to answer this.
Answer:
At (-2,0) gradient is -4 ; At (2,0) gradient is 4
Step-by-step explanation:
For this problem, we simply need to take the derivative of the function and evaluate when y = 0 (when crossing the x-axis).
y = x^2 - 4
y' = 2x
The function y = x^2 - 4 cross the x-axis when:
y = x^2 - 4
0 = x^2 - 4
4 = x^2
2 +/- = x
Hence, this curve crosses the x-axis twice, once at (-2,0) and again at (2,0).
The gradient at these points are as follows:
y' = 2(-2) = -4
y' = 2(2) = 4
Cheers.
