How many square units are in the area of the triangle whose vertices are the x and y intercepts of the curve y = (x-3)^2 (x+2)?
1 answer:
The intercepts can be read from the equation as (-2, 0), (0, 18), (3, 0). These define a triangle with a base of 5 and an altitude of 18, so the area is
... A = (1/2)·5·18 = 45 . . . . square units
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The x-intercepts are those values of x that make one or another of the factors be zero. (x-3) is zero when x=3; (x+2) is zero when x=-2.
The y-intercept is the value when x=0. That is (-3)²·(2) = 9·2 = 18.
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Answer:
Using Geometry to answer the question would be the simplest:
Step-by-step explanation:
Remembering the formula for the area of a triangle which is . One can then tackle the question by doing the following:
Step 1 Find the y-intercepts
The y-intercepts are found by substituting in .
Which gives you this when you plug it into both equations:
So the y-intercepts for the graphs are , and
respectively.
Now one has to use elimination to solve the problems by adding up the equations we get:
Now to solve for the y component substitute:
Therefore, the graphs intersect at the following:
Now we have our triangle which is accompanied by the graph.
now to solve it we must figure out how long the base is:
The height must also be accounted for which is the following:
Now the formula can be used:
