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.
B) would be tha answer :)
Answer:
A circle is shown. Secants P N and L N intersect at point N outside of the circle. Secant P N intersects the circle at point Q and secant L N intersects the circle at point M. The length of P N is 32, the length of Q N is x, the length of L M is 22, and the length of M N is 14.
In the diagram, the length of the external portion of the secant segment PN is <u>X</u>
The length of the entire secant segment LN is <u>36</u>.
The value of x is <u>15.74</u>
Step-by-step explanation:
Snap
Jona_Fl16
