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.
Answer:
Zeroes: None
Domain: All real numbers
Maximum: None
Step-by-step explanation:
EDGE PRECAL 2020
then its 658.74 or 6.5874
∑x = 1 + 2 + 3 + 4 + 5 + 6 = 21
∑y = 8 + 3 + 0 + 1 + 2 + 1 = 15
∑x^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91
∑y^2 = 64 + 9 + 0 + 1 + 4 + 1 = 79
∑xy = 8 + 6 + 0 + 4 + 10 + 6 = 34
r
= (n∑xy - ∑x∑y)/(sqrt(n∑x^2 - (∑x)^2)*sqrt(n∑y^2 - (∑y)^2)) = (6(34) -
21(15))/(sqrt(6(91) - (21)^2)*sqrt(6(79) - (15)^2)) = (204 -
315)/(sqrt(546 - 441)*sqrt(474 - 225)) = -111/(sqrt(105)*sqrt(249)) =
-111/(10.25*15.78) = -111/161.7 = -0.68
Use Ga_thMath(u) (brainly doesn't allow me to type it) To use the app u need to take a pic of the problem and then it will process it and you'll get ur answer ASAP(most of the time). Many questions have been asked before so search it on brainly.
