Find line tangent to y=xcosx at point (pi, -pi)

1 answer:

5 0

1) slope of the line = derivative of the of the function at the given point.

y' = cosx - x cosx

Evaluate for x = pi => y' = cos(pi) - pi*cos(pi) = -1 - (pi*cos(pi)) = -1 + pi

Then slope = -1 + pi

Given point: (pi, -pi)

Equation:

y - (-pi) = [-1+ pi] (x - pi)

y + pi = -x +pi + xpi -pi^2

y = (pi-1)x - pi^2 .... this is the answer

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Rate of change y=5x+10

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Answer:

Step-by-step explanation:

The equation $y=5x+10$ is given in slope-intercept form. Slope-intercept form is a way of writing the equation of a line in the format $y=mx+b$ , such that the coefficient of the parameter (x) is the slope, and constant (b) is the y-intercept. What that essentially means is,

$y = mx + b\\\\m=slope\\b = y-intercept.$

The slope of a line is its rate of change, and the y-intercept is where the graph of the line intersects the y-axis. One can apply this knowledge here by saying that, the equation $y=5x+10$ is given in the slope-intercept form, and the coefficient ( $m$ ) equals ( $5$ ), therefore the slope is ( $5$ ).