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The equation of the curve which passes through the point (1,2) and has the slope 3x -4 and the point of (x, y) is (a) 2y = 3x^2-8x+9 (b) y = 6x^2-8x+9 (c) y=x^2-8x+9 (d) 2y= 3x^2-8x+c?
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The equation of the curve which passes through the point (1,2) and has...
Given information:
- Point (1,2) lies on the curve.
- The slope of the curve is given by 3x - 4.
Approach:
To find the equation of the curve, we need to integrate the given slope function with respect to x. The resulting equation will represent the curve.
Solution:
Step 1: Integrating the slope function
The slope function is given by 3x - 4. To integrate this function, we need to find the antiderivative of the function.
∫(3x - 4) dx = ∫3x dx - ∫4 dx = 3∫x dx - 4∫1 dx
Integrating, we get:
= 3(x^2/2) - 4(x) + C
Where C is the constant of integration.
Step 2: Finding the constant of integration
To find the constant of integration (C), we can use the fact that the curve passes through the point (1,2). Substituting x = 1 and y = 2 into the equation, we can solve for C.
2 = 3(1^2/2) - 4(1) + C
2 = 3/2 - 4 + C
2 = -5/2 + C
C = 2 + 5/2
C = 9/2
Step 3: Writing the equation of the curve
Now that we have the constant of integration, we can write the equation of the curve.
y = 3(x^2/2) - 4(x) + 9/2
y = 3/2 * x^2 - 4x + 9/2
Final Answer:
The equation of the curve that passes through the point (1,2) with a slope of 3x - 4 is:
y = 3/2 * x^2 - 4x + 9/2
Therefore, the correct option is (c) y = x^2 - 8x + 9.
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The equation of the curve which passes through the point (1,2) and has the slope 3x -4 and the point of (x, y) is (a) 2y = 3x^2-8x+9 (b) y = 6x^2-8x+9 (c) y=x^2-8x+9 (d) 2y= 3x^2-8x+c?
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