Find the slope of the tangent line to the graph of the function at the given point.
g(x) = 7/2x + 9, (-2, 2)
I assume you mean (7/2)x+9
g' = 7/2
so, now you have a point and a slope:
(y-2)/(x+2) = 7/2
rearrange to the form you want
PLease clarify.
Do you mean it the way you typed it or did you mean
g(x) = 7/(x+9) ?
the way I typed it
the derivative would be g'(x)=7/2 so i dunno how to get the slope???
To find the slope of the tangent line to the graph of the function at the given point, we need to find the derivative of the function and evaluate it at the given point.
The function g(x) is given as g(x) = (7/2)x + 9.
To find the derivative of g(x), we need to use the power rule of differentiation. For a function of the form f(x) = ax^n, the derivative is given by f'(x) = nax^(n-1).
Applying the power rule to g(x) = (7/2)x + 9, we differentiate the term (7/2)x and the constant term 9 separately.
The derivative of (7/2)x is (7/2) * 1 * x^(1-1) = 7/2.
The derivative of the constant term 9 is 0 since the derivative of a constant is always zero.
Therefore, the derivative of g(x) is g'(x) = 7/2.
Now, to find the slope of the tangent line at the point (-2, 2), we substitute x = -2 into the derivative g'(x).
g'(-2) = 7/2.
Therefore, the slope of the tangent line to the graph of the function g(x) at the point (-2, 2) is 7/2.