Respuesta :
The value for h(-1) is 4
The value for h'(-1) is 1
Given the tangent line to the graph of y=h(x) at the point (-1,4) passes through (3,6), to get h(-1) and h'(-1), we should calculate the following:
Gradient of the tangent line:
Gradient=change in y axis/change in x axis=Δy/Δx
Using the x and y coordinates (-1,4) and (3,6)
Gradient=(6-4)/(3--1)
=2/4
=1/2
Gradient=1/2 or 0.5
Equation of the tangent line:
This should be in the form of y=mx+c, where m is the gradient of the line
We use the gradient and one of the x and y coordinates get the equation of the tangent line
Gradient=change in y axis/change in x axis=Δy/Δx, and gradient was found as 1/2
Δy/Δx=1/2
(y-4)/(x--1)=1/2
(y-4)/(x+1)=1/2
2y-8=x+1
y=(1/2)x+4.5
Equation of the tangent line is y=(1/2)x+4.5 or y=0.5x+4.5
And y=h(x)
So, h(x)=0.5x+4.5
Calculate h(-1), here x=-1:
h(x)=0.5x+4.5
h(-1)=(0.5*-1)+4.5
=4
h(-1)=4
Calculate h'(-1), x=-1:
Here we get the derivative of the equation of the tangent line h(x)=0.5x+4.5,
h'(x)=0.5x^0+0
h'(-1)=1+0
h'(-1)=1
Learn more about equation of tangents here:
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