A parabola with the vertex at (1.5, 12.5) inscribes a segment of length 5 on x-axis. What is the area of the triangle formed by
points of intersection of parabola with the coordinate axes.
1 answer:
Answer:
20 units²
Step-by-step explanation:
The x-intercepts are symmetrically located around the x-coordinate of the vertex, so are at
1.5 ± 5/2 = {-1, 4}
Using one of these we can find the unknown parameter "a" in the parabola's equation (in vertex form) ...
0 = a(4 -1.5)² +12.5
0 = 6.25a +12.5 . . . . . simplify
0 = a +2 . . . . . . . . . . . divide by 6.25
-2 = a
Then the standard-form equation of the parabola is ...
y = -2(x -1.5)² +12.5 = -2(x² -3x +2.25) +12.5
y = -2x² +6x +8
This tells us the y-intercept is 8. Then the relevant triangle has a base of 5 units and a height of 8. Its area is given by the formula ...
A = (1/2)bh = (1/2)(5)(8) = 20 . . . . units²
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well, about A and D, I just plugged the values on the slope formula of
for A the values are 8.01 and 8.0, so indeed those "slopes" are close.
for D the values are -2.25 and 8.0, so no dice on that one.
for B, let's check the y-intercept for g(x), by setting x = 0, we end up g(0) = 0³-4(0)+1, which gives us g(0) = 1.
checking L(x) y-intercept, well, L(x) is in slope-intercept form, thus the +1 sticking out on the far right is the y-intercept, so, dice.
for C, well, the slope if L(x) is 8, since it's in slope-intercept form, the derivative of g(x) is g'(x) = 3x² - 4, and thus g'(0) = -4, so no dice.
for E, do they intercept at (2,1)? well, come on now, L(x) is a tangent line to g(x), so that's a must for a tangent.
for F, we know the slope of the line L(x) is 8, is g'(2) = 8? let's check
recall that g'(x) = 3x² - 4, so g'(2) = 3(2)² - 4, meaning g'(2) = 8, so, dice.
Given that the volume of the aquarium is 20m^3.
Volume = Area of Base x height
Area of Base = Volume / height = 20/h
Given that the aquarium has a square base.
Area of square = l^2
Thus, the length of the base of the aquarium is
The frame is to cover 8 sides with the length equal to the length of the base and 4 sides with the length of the height.
Thus, the total perimeter of the frame is given by
Area of the four side faces of the aquarium is 4 times the length of the base times the height =
Total area to be covered by grass is the base and the four side faces and is given by
Cost of the entire metal frame =
Cost of the entire grass =
Therefore, total cost in terms of the height, h, is given by
Volume = Area of Base x height
Area of Base = Volume / height = 20/h
Given that the aquarium has a square base.
Area of square = l^2
Thus, the length of the base of the aquarium is
The frame is to cover 8 sides with the length equal to the length of the base and 4 sides with the length of the height.
Thus, the total perimeter of the frame is given by
Area of the four side faces of the aquarium is 4 times the length of the base times the height =
Total area to be covered by grass is the base and the four side faces and is given by
Cost of the entire metal frame =
Cost of the entire grass =
Therefore, total cost in terms of the height, h, is given by