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- Problem 1 Find the equation of the line which passes through the points (2, 3) and (โ 2 , 1). Sketch the graph of the line.
Solution 1. The slope of the line is given by
โy โx
Hence, the equation of the line in point-slope form is
(2) y โ 3 =
ยท (x โ 2).
We rewrite the equation in slope-intercept form:
y โ 3 = x 2
y = x 2
Hence, the y-intercept of the line is 2. Setting y = 0 we obtain
0 = x 2
x 2 โ4 = x.
Hence, the x-intercept is โ4. Let f (x) = x 2 + 2. We sketch the graph of f in figure 1.
- Problem 2 Sketch the graph of y^2 โ 4 y + 10 + x^2 โ 9 x = 0. 1
y=f(x)
โ
2 x
y
Figure 1. Illustration for problem 1
Solution 2. We complete the square in both the y and x variables:
y^2 โ 4 y + 10 + x^2 โ 9 x = 0
(y โ 2)^2 โ 4 + 10 +
x โ
(y โ 2)^2 +
x โ
(y โ 2)^2 +
x โ
(y โ 2)^2 +
x โ
(y โ 2)^2 +
x โ
Hence, the graph of our equation is the circle centered at
with
radius
โ 57
(2, 4.5)
x
y
ฮฑ ฮฒ
F(x,y)=
Figure 2. Illustration for problem 2
Notice that
f (2 + h) โ f (2) = (2 + h)^2 +
2 + h
= 2^2 + 4h + h^2 โ 22 +
2 + h
= 4h + h^2 + 2 โ (2 + h) (2 + h) = 4h + h^2 โ h (2 + h) f (2 + h) โ f (2) = h
4 + h โ
(2 + h)
Dividing through by h we obtain:
(13) f (2 + h) โ f (2) h = 4 + h โ
(2 + h)
Sending h to 0, we see that
(14) f โฒ(2) = 4 โ
- Problem 5 What is the equation of the tangent line to the unit circle at the point
โ^1 2 ,^ โ^1 2
Solution 5. The slope of the line segment connecting
โ^1 2 ,^ โ^1 2
to the origin (which is the center of the circle) is 1. Hence, the slope of the tangent line to the circle at our point is โ1. In point-slope form, the equation of the tangent line is
(15) y โ
x โ
Solving for y, we obtain
(16) y = โx +
2 โ x.
- Problem 6 Compute
(17) lim xโ 3 x^2 โ 9 x โ 3
Solution 6. Notice that for x 6 = 3,
(18) x^2 โ 9 x โ 3 = x + 3.
Hence,
(19) (^) xlimโ 3 x^2 โ 9 x โ 3 = lim xโ 3 (x + 3) = 6.
