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The slope of the tangent line to a curve is given by [tex]f^{\prime}(x) = 2x^2 + 5x - 9[/tex]. If the point [tex](0,6)[/tex] is on the curve, find an equation of the curve.

[tex]f(x) = [/tex]


Sagot :

To find the equation of the curve given the derivative [tex]\( f'(x) = 2x^2 + 5x - 9 \)[/tex] and the point [tex]\((0, 6)\)[/tex] on the curve, follow these steps:

1. Find the original function [tex]\( f(x) \)[/tex] by integrating the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f(x) = \int (2x^2 + 5x - 9) \, dx \][/tex]

2. Compute the integral:
[tex]\[ \int (2x^2 + 5x - 9) \, dx = \int 2x^2 \, dx + \int 5x \, dx - \int 9 \, dx \][/tex]
Breaking it down:
[tex]\[ \int 2x^2 \, dx = \frac{2}{3}x^3 \][/tex]
[tex]\[ \int 5x \, dx = \frac{5}{2}x^2 \][/tex]
[tex]\[ \int 9 \, dx = 9x \][/tex]
So, combining these results:
[tex]\[ f(x) = \frac{2}{3}x^3 + \frac{5}{2}x^2 - 9x + C \][/tex]
where [tex]\( C \)[/tex] is the integration constant.

3. Determine the value of the constant [tex]\( C \)[/tex] using the given point [tex]\((0, 6)\)[/tex] on the curve:
Substitute [tex]\( x = 0 \)[/tex] and [tex]\( f(0) = 6 \)[/tex] into the equation:
[tex]\[ 6 = \frac{2}{3}(0)^3 + \frac{5}{2}(0)^2 - 9(0) + C \][/tex]
Simplifying this, we get:
[tex]\[ 6 = C \][/tex]

4. Write the final equation of the curve:
Replace [tex]\( C \)[/tex] with 6 in the equation:
[tex]\[ f(x) = \frac{2}{3}x^3 + \frac{5}{2}x^2 - 9x + 6 \][/tex]

Therefore, the equation of the curve is:
[tex]\[ f(x) = \frac{2}{3}x^3 + \frac{5}{2}x^2 - 9x + 6 \][/tex]