How do you graph the parabola y= (x + 3)^2 - 2y=(x+3)2−2 using vertex, intercepts and additional points?

1 Answer
Nallasivam V
Jul 10, 2016

Refer explanation section

Explanation:

The given quadratic equation is in the vertex form

y=(x-3)^2-2y=(x−3)2−2

Hence the vertex is (3, -2)(3,−2)

(3, -2)(3,−2)This is one of the points on the curve.

x=-3x=−3 is the minimum point on the curve. Hence to graph the curve, we take two point to the left of x=3x=3 and two point to its right.

Right side points -

At x=5; y=(5-3)^2-2=4-2=2x=5;y=(5−3)2−2=4−2=2

(5,2)(5,2)

At x=4; y=(4-3)^2-2=1-2=-1x=4;y=(4−3)2−2=1−2=−1

(-4,-1)(−4,−1)

(3, -2)(3,−2)

Left side points.

At x=2; y=(2-3)^2-2=1-2=-1x=2;y=(2−3)2−2=1−2=−1

(-2, -1)(−2,−1)

At x=1; y=(1-3)^2-2=4-2=2x=1;y=(1−3)2−2=4−2=2

(-1,2)(−1,2)

Plot the points
(5, 2), (4, -1), (3, -2), (2, 1), (-1, 2)(5,2),(4,−1),(3,−2),(2,1),(−1,2)

Table developed in Excel for the equation y=(x-3)^2-2y=(x−3)2−2

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You will get the graph.

Graph developed in Excel

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