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15 Jul 2020, 08:24
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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63% (02:14) correct
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What is the area of the triangle formed by connecting the points where the parabola \(y=2x^2-7x-4\) intersects the \(x\) and \(y\)-axes?
A) 6
B) 7
C) 8
D) 9
E) 10
A) 6
B) 7
C) 8
D) 9
E) 10
15 Jul 2020, 08:55
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The roots of the equation 2\(x^2\) - 7x - 4 = 0 can be found by splitting the middle term into -8x and x
2\(x^2\) - 8x + x - 4 = 0
2x (x - 4) + 1(x - 4)
(2x + 1)(x - 4) = 0
x = -\(\frac{1}{2}\) and 4 are the roots. The parabola cuts the x axis at these points (-0.5,0) and (4,0)
The y intercept can be found by putting x = 0 in the equation, 2\(x^2\) - 7x - 4 = 0
We get y = -4. The point at which the parabola cuts the y axis is (0, -4)
Refer to the figure below.
Area of the triangle = \(\frac{1}{2}\) * base * height
The area of \(\triangle{ABC}\) = area of \(\triangle{AOB}\) + area of \(\triangle{OBC}\)
= \(\frac{1}{2}\) * 0.5 * 4 + \(\frac{1}{2}\) * 4 * 4 = 1 + 8 = 9
Option D
Arun Kumar
2\(x^2\) - 8x + x - 4 = 0
2x (x - 4) + 1(x - 4)
(2x + 1)(x - 4) = 0
x = -\(\frac{1}{2}\) and 4 are the roots. The parabola cuts the x axis at these points (-0.5,0) and (4,0)
The y intercept can be found by putting x = 0 in the equation, 2\(x^2\) - 7x - 4 = 0
We get y = -4. The point at which the parabola cuts the y axis is (0, -4)
Refer to the figure below.
Area of the triangle = \(\frac{1}{2}\) * base * height
The area of \(\triangle{ABC}\) = area of \(\triangle{AOB}\) + area of \(\triangle{OBC}\)
= \(\frac{1}{2}\) * 0.5 * 4 + \(\frac{1}{2}\) * 4 * 4 = 1 + 8 = 9
Option D
Arun Kumar
Attachments
Parabola.jpg [ 382.57 KiB | Viewed 2771 times ]
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15 Jul 2020, 09:03
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the equation y = 2x^2-7x-4 intersect both X and Y axis
When it intersect Y axis, X is equal to zero
y = 2x^2-7x-4
y = 2(0)^2-7(0)-4
y = -4
When it intersect x axis, y is equal to zero
y = 2x^2-7x-4
0 = 2x^2-7x-4
0 = 2x^2-8x+x-4
0 = 2x(x-4)+1(x-4)
0 = (x-4)(2x+1)
x = 4 and -1/2
plotting points on graph, we get triangle ABC
Area of triangle BOC = 1/2*4*4 = 8
Area of triangle AOC = 1/2*1/2*4 = 1
Area of ABC = Area of BOC + Area of AOC
Area of ABC = 8+1
Area of ABC = 9
Answer is D
When it intersect Y axis, X is equal to zero
y = 2x^2-7x-4
y = 2(0)^2-7(0)-4
y = -4
When it intersect x axis, y is equal to zero
y = 2x^2-7x-4
0 = 2x^2-7x-4
0 = 2x^2-8x+x-4
0 = 2x(x-4)+1(x-4)
0 = (x-4)(2x+1)
x = 4 and -1/2
plotting points on graph, we get triangle ABC
Area of triangle BOC = 1/2*4*4 = 8
Area of triangle AOC = 1/2*1/2*4 = 1
Area of ABC = Area of BOC + Area of AOC
Area of ABC = 8+1
Area of ABC = 9
Answer is D
Attachments
WhatsApp Image 2020-07-15 at 21.21.35 (1).jpeg [ 119.92 KiB | Viewed 2685 times ]
