Help~ Don't remember what the point of tengency is.

[baby_garfield]

Here is the problem.

f(x)=x^3-18x^2=105x-146 and 2 is a real zero.

now, through the point (2,0), draw a line that is tangent to the graph at a point to the right of the low point at x=7. find the slope of the tangent line using x=2 and the point of tangency.

and here is the graph.

 

[HallsofIvy]

f(x)=x^3-18x^2=105x-146 and 2 is a real zero.

I assume that is f(x)= x³- 18x²+ 105 x- 146.
You want to find a line that passes through (2, 0) and is tangent to the graph at some (x,y) where x> 7.

Okay: any line through (2, 0) can be written as y= m(x-2) where m is the slope.

If the line is tangent to the graph at, say x₁ , then it must satisfy the equation at that point: we must have m(x₁- 2)= x₁³- 18x₁²+ 105 x₁- 146.
It also must have the same slope there so m= f '(x₁)= 3x₁²- 36x₁+ 105.

Solve those two equations for m and x₁.

 

[tacman]

The point of tangency is the point where a line touches a curve at only one point, without crossing over or intersecting the curve. In this problem, the point (2,0) is a real zero of the function f(x)=x^3-18x^2-105x+146. This means that when x=2, the value of the function is equal to 0.

To find the slope of the tangent line at the point (2,0), we can use the formula for slope, which is rise over run. In other words, the slope is equal to the change in y over the change in x.

To find the change in y, we can use the given point (2,0) and the point of tangency (x,y). The change in y is equal to y-0, or simply y.

To find the change in x, we can use the given point (2,0) and the point of tangency (x,y). The change in x is equal to x-2.

Now, we can set up the equation for slope using these values:

slope = (y-0)/(x-2)

To solve for the slope, we need to find the coordinates of the point of tangency. From the graph, we can see that the low point of the curve is at x=7. This means that the point of tangency must be to the right of x=7.

To find the coordinates of the point of tangency, we can plug in x=7 into the original function:

f(7)=7^3-18(7)^2-105(7)+146=0

This means that the point of tangency is (7,0).

Now, we can plug in the values into the equation for slope:

slope = (0-0)/(7-2) = 0

Therefore, the slope of the tangent line at the point (2,0) is 0. This means that the tangent line is horizontal at this point.

To draw the tangent line, we can plot the points (2,0) and (7,0) and draw a straight line connecting them. This line will be tangent to the curve at the point (7,0).

In summary, the point of tangency is the point where a line touches a curve at only one point without crossing over or intersect