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The area of the isosceles triangle is 64 sq units.
<u>Solution:</u>
Part 1: x-intercepts
The x-intercepts occur at the points on the function where y=0
So, we need to solve
The left side factors fairly easily into:
So solution occur when
and
So the x-intercepts are at (0,6) and (0,−2)
Part 2: vertex of the parabola
The vertex of a simple quadratic parabola occurs when the derivative of the quadratic is equal to 0.
The derivative of the given quadratic is
By observation, this is equal to 0 when x=2
When x=2 the original equation becomes
Therefore the vertex of this parabola is at (2,−16)
The endpoints of the base of the isosceles triangle are (-6, 0) and (2, 0)
so its base is 8
The height of the triangle reaches from the midpoint of the base (-2, 0) and the vertex (2, -16)
so its height is 16
The area is
Answer:
(3,-2)
Step-by-step explanation:
Part A:
1) <em>Isolate x in x - y = 5</em>
x = y + 5
2) <em>Substitute that x into 4x+y=10</em>
4(y + 5) + y = 10
4y + 20 + y = 10
5y = 10-20
5y = -10
y = -2
3) Use y to solve for x:
x = -2 + 5
x = 3
Part B:
<em>Verify using results from Part A</em>
3 - (-2) = 5
4(3) + (-2) = 12 - 2 = 10
Part C:
<em>Just graph the equations and see where they intersect, which is exactly (3,-2). We got that in Part A and B</em>
