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10) Find the equation of tangent for the equation \( y=x^{2}-5 \sqrt{x}+3 \) at \( x=9 \)

Ask by Delgado Maxwell. in Malaysia
Mar 17,2025

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The equation of the tangent to the curve \( y = x^2 - 5 \sqrt{x} + 3 \) at \( x = 9 \) is: \[ y - 69 = \frac{103}{6}(x - 9). \]

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To find the equation of the tangent line to the curve \( y = x^2 - 5\sqrt{x} + 3 \) at \( x = 9 \), we first need to compute the derivative, which gives the slope of the tangent. The derivative is \( y' = 2x - \frac{5}{2\sqrt{x}} \). Evaluating this at \( x = 9 \) yields a slope of \( y'(9) = 18 - \frac{5}{6} = \frac{103}{6} \). Next, we calculate the y-coordinate of the point on the curve at \( x = 9 \): \( y(9) = 9^2 - 5\sqrt{9} + 3 = 81 - 15 + 3 = 69 \). Now, we can use the point-slope form of a line, which is \( y - y_1 = m(x - x_1) \). Here, \( m = \frac{103}{6} \), \( x_1 = 9 \), and \( y_1 = 69 \). Thus, the equation of the tangent line is: \[ y - 69 = \frac{103}{6}(x - 9) \]. You can rearrange this into the slope-intercept form if desired! You know, this process not only applies to parabolas like this one but also to all sorts of curves. Understanding tangents can help us in physics, engineering, and even game development, where understanding motion trajectories is key! Interestingly, the concept of tangents has been explored for centuries. Ancient Greeks, like Archimedes, had notions similar to calculus, using lines that just touched curves. Fast forward to the 17th century, and along comes Newton and Leibniz, who formalized these ideas into what we now call calculus. So, the equation of that tangent is a product of a long mathematical journey!

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