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In Exercises 9-12, at the indicated point find

(a) the slope of the curve,
(b) an equation of the tangent, and
(c) an equation of the normal.
(d) Then draw a graph of the curve, tangent line, and normal
line in the same square viewing window.
9. y=x² at x = -2

User Singingsingh
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2 Answers

12 votes
12 votes

Answer:

y'(-1*2 )= -4

Explanation:

y'=2x

y'(-1*2) = 2 ⋅ (-2)

-4

y'(-1*2) = -4

User Shihabudheen K M
by
2.7k points
16 votes
16 votes

Answer:

(a) slope: -4

(b) tangent: y -4 = -4(x +2)

(c) normal: y -4 = 1/4(x +2)

(d) graph: see attached

Explanation:

You want the slope of the curve, and equations for the tangent and normal lines at x = -2 when y = x².

(a) Slope

The slope of the curve is given by the derivative.

y' = 2x

At x = -2, the slope is

y' = 2(-2) = -4

The slope of the curve at x=-2 is -4.

(b) Equation of the tangent

The value of y at x=-2 is ...

y = (-2)² = 4

The point-slope equation of the line with slope -4 through point (-2, 4) is ...

y -k = m(x -h) . . . . . . line with slope m through point (h, k)

y -4 = -4(x +2) . . . . . . line with slope -4 through point (-2, 4)

(c) Equation of the normal

The normal has the opposite reciprocal slope at the point of tangency.

y -4 = 1/4(x +2) . . . . . . . line with slope 1/4 through point (-2, 4)

(d) Graph

See the attachment for a graph.

In Exercises 9-12, at the indicated point find (a) the slope of the curve, (b) an-example-1
User Zyberzero
by
3.4k points