Derivatives concept: Equation of the secant line and tangent to a curve.
Let the function and its graph be:

(In both graphs (activity A and B) all the corresponding development must be carried out to arrive at the requested equation)

9514 1404 393

  A. y = -2x +13

  B. y = 8x -7

Step-by-step explanation:

A. We can read the y-intercept of the secant line from the graph. It is 13.

The slope can also be read from the graph, but we choose to use the slope formula:

  m = (y2 -y1)/(x2 -x1)

  m = (19 -9)/(-3 -2) = 10/-5 = -2

Then the slope-intercept formula for the line is ...

  y = mx + b . . . . . . line of slope m and y-intercept b

  y = -2x +13

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B. The vertex of the given parabola is (0, 1). We notice that when x=1 (1 unit right of the vertex), y = 3 (2 units up from the vertex). This tells us the vertical scale factor of the parabola is 2. That means the vertex form equation is ...

  y = a(x -h)^2 +k . . . . . . . . vertex (h, k), scale factor 'a'

  y = 2(x-0)^2 +1 . . . . . . . use known values for (h, k)

  y = 2x^2 +1

The derivative of this is ...

  y' = 4x

So, at x=2, the given point A, the slope of the tangent line is ...

  m = y' = 4(2) = 8

We have a point and the slope, so we can write the point-slope form of the equation for the tangent line:

  y -9 = 8(x -2)

Rearranging to slope-intercept form, this is ...

  y = 8x -7

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Additional comment

You can also read the slope of the tangent line from the graph. The line also goes through the point (1, 1), so has a rise of 8 for a run of 1. The y-intercept can be found from ...

  b = y -mx = 9 -8(2) = -7

This lets you write the equation of the tangent line directly from the graph.

That is, the parameters of both lines can be read from the graph, so there is very little "development" required.

99

step-by-step explanation:

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50