Y=(-x*x)-4x+3
y=-(x^2+4x-3)
4+y=-(x^2+4x+4)+3
y+4=-(x-2)^2.) +3
-4. -4
y=-((x-2)^2)-1
Answers:
Vertex: (2,-1)
AOS:x = 2
Domain: All Real Numbers
Range:[-1,Infinity)
Formula for slope: y2-y1/x2-x1
Plug in values.
3-8/-4-(-7)= -5/3 = slope
42-1/36-12=41/24=slope
Answer:
Choice B is correct.
Explanation:
The given equation is -r+5(r+2)=-10
Put r=-5 into the above equation
-(-5)+5(-5+2)=-10
5+5(-3)=-10
5-15=-10
-10=-10
Right and left side of above equation is equal by putting r=-5 so it is true statement.
Answer:
(f⁻¹)'(b) = 1/f'(f⁻¹(b)) = 1/f'(a)
Step-by-step explanation:
The function f⁻¹(x) is the reflection of the function f(x) across the line y=x. Every point (a, b) that is on the graph of f(x) is reflected to be a point (b, a) on the graph of f⁻¹(x).
Any line with slope m reflected across the line y=x will have slope 1/m. (x and y are interchanged, so m=∆y/∆x becomes ∆x/∆y=1/m) Since f'(x) is the slope of the tangent line at (x, f(x)), 1/f'(x) will be the slope of the tangent line at (f(x), x).
Replacing x with f⁻¹(x) in the above relation, you get ...
... (f⁻¹)'(x) = 1/f'(f⁻¹(x)) will be the slope at (x, f⁻¹(x))
Putting your given values in this relation, you get
... (f⁻¹)'(b) = 1/f'(f⁻¹(b)) = 1/f'(a)
The function is nonlinear.
