1. use the definition mtan=(f(x)-f(x))/(x-a) to find the SLOPE of the line tangent to the graph of f at P.
2. Determine an equation of the tangent line at P.
3. Given 1 & 2, how would I plot the graph of f and the tangent line at P if : f(x)=x^2 +4, P(4,20)
first of all, if you are trying to define the definition of the derivative, it should be
mtan=Lim (f(x+a)-f(x))/(a) , as a --->∞
Furthermore, I will assume that f(x) = x^2 + 4, you only state that at the very end
f(x+a) = (x+a)^2 + 4
then (f(x+a)-f(x))/(a) = (x^2 + 2ax + a^2 + 4) - x^2 - 4)/a)
slope of tangent = lim (x^2 + 2ax + a^2 + 4 - x^2 - 4)/a as a --->∞
= lim (2ax + a^2)/a , as a --->∞
= lim a(2x + a)/a , as a --->∞
= lim 2x + a , as a --->∞
= 2x
now at the point (4,20)
the slope of the tangent is 2(4) or 8
and the tangent equation is
y-20 = 8(x-4)
y = 8x - 12
check with Wolfram:
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+%2B+4,+y+%3D+8x+-+12
a --->∞ should probably be a --->0
Scott, of course you are right, my silly error.
Fortunately, my calculations match the a --->0 conditions
To find the slope of the line tangent to the graph of f at point P, you can use the definition of the derivative.
1. Use the definition mtan = (f(x) - f(a)) / (x - a), where mtan represents the slope of the tangent line, f(x) is the function value at some point x, f(a) is the function value at the point of tangency a, and (x - a) is the difference between the x-coordinate of the point of tangency and the x-coordinate of the point on the tangent line. In this case, we have f(x) = x^2 + 4 and the point of tangency P(4, 20).
Plug in the values into the formula:
mtan = (f(x) - f(a)) / (x - a)
= (f(x) - f(4)) / (x - 4)
= (x^2 + 4 - 20) / (x - 4)
Simplifying the equation further, we have:
mtan = (x^2 - 16) / (x - 4)
mtan = (x - 4)(x + 4) / (x - 4)
mtan = x + 4
So, the slope of the tangent line at P is x + 4.
2. To determine the equation of the tangent line at P, we need to use the point-slope form of the equation of a line. The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) represents a point on the line and m is the slope of the line. In this case, we know that the point of tangency is P(4, 20) and the slope of the tangent line is x + 4.
Using the point-slope formula, we substitute the values:
y - 20 = (x + 4)(x - 4)
Simplifying, we expand the terms:
y - 20 = x^2 - 16
Now, rearrange the equation to solve for y:
y = x^2 - 16 + 20
y = x^2 + 4
So, the equation of the tangent line at P is y = x^2 + 4.
3. To plot the graph of f(x) = x^2 + 4 and the tangent line at P(4, 20), you can follow these steps:
- First, plot the graph of f(x) = x^2 + 4. This is a parabola that opens upwards.
- Choose some x-values and calculate the corresponding y-values based on the equation f(x) = x^2 + 4.
- Plot the points (x, f(x)) on the graph.
- Connect the points to form the parabolic curve.
- Next, plot the point P(4, 20) on the graph. This point represents the point of tangency.
- Finally, plot the tangent line at P.
- Use the equation of the tangent line, y = x^2 + 4, and select a few x-values.
- Calculate the corresponding y-values based on the equation.
- Plot the points (x, y) on the graph.
- Connect the points to form the tangent line passing through point P(4, 20).
Make sure to label the parabola and the tangent line to distinguish between them.