Find (a) the slope of the curve at the given point p, and (b) an equation of the tangent line at p. y=−2−9x2; p(2,−38)
1 answer:
Answer:
a) -36
b) y = -36x +34
Step-by-step explanation:
a) The slope is found using the derivative of the function.
f'(x) = dy/dx = 0 -9(2x) = -18x
Then at x=2, the slope is ...
f'(2) = -18·2 = -36
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b) The point-slope form of the equation can be written as ...
y = -36(x -2) -38
Simplifying to slope-intercept form, we get ...
y = -36x +34
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Answer:
Slope of the base segment line is zero, hence base segment is horizontal
Slope of the segment that joins the vertex angle to the midpoint of its base is undefined hence the line is vertical
Therefore, angle between base segment line and the segment line from the vertex angle to the midpoint is perpendicular
Therefore, the segment that joins the vertex angle to an isosceles triangle to the midpoint of its base is perpendicular to the base
Step-by-step explanation:
Here we prove the required relation as follows;
Let the isosceles be ABC
The coordinates of the points are
C = (0, 0) (Vertex)
A = (-a, b)
C = (a, b)
P = (0, b) (Midpoint of base)
Therefore, the gradient or slope of the base AC is presented as follows;
Hence, segment AC is vertical
Hence, segment CP is vertical
Therefore, the segment that joins the vertex angle to an isosceles triangle to the midpoint of its base is perpendicular to the base.