James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A22
appendix c Graphs of Second-Degree Equations
and translate it (shift it) so that its center is the point sh, kd, then its equation becomes
5
sx 2 hd 2
a 2
1
sy 2 kd2
b 2 − 1
(See Figure 14.)
y
(x-h)@ (y-k)@
+ =1
a@ b@
b
(h, k)
≈ ¥
a@ + b@
=1 a
(x, y)
b
k
FIGURE 14
(0, 0)
a
(x-h, y-k)
h
x
Notice that in shifting the ellipse, we replaced x by x 2 h and y by y 2 k in Equation
4 to obtain Equation 5. We use the same procedure to shift the parabola y − ax 2 so that
its vertex (the origin) becomes the point sh, kd as in Figure 15. Replacing x by x 2 h and
y by y 2 k, we see that the new equation is
y 2 k − asx 2 hd 2 or y − asx 2 hd 2 1 k
y
y=a(x-h)@+k
y=a≈
(h, k)
FIGURE 15
0 x
EXAMPLE 7 Sketch the graph of the equation y − 2x 2 2 4x 1 1.
SOLUTIOn First we complete the square:
y − 2sx 2 2 2xd 1 1 − 2sx 2 1d 2 2 1
In this form we see that the equation represents the parabola obtained by shifting
y − 2x 2 so that its vertex is at the point s1, 21d. The graph is sketched in Figure 16.
y
1
FIGURE 16
y − 2x 2 2 4x 1 1
0
1 2 3 x
(1, _1)
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