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$\begin{aligned} y &=(x-5)^2+1\ \dfrac{y-5}{3}&=15 \end{aligned}$

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Find equations of tangent lines using limits

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Q.

\begin{aligned} y &=(x-5)^2+1\ \dfrac{y-5}{3}&=15 \end{aligned}

Solve for y: First, let's solve the second equation for y. $\newline$ Given: $\dfrac{y-5}{3} = 15$ $\newline$ To isolate y, we multiply both sides by $3$ . $\newline$ $\dfrac{y-5}{3} \times 3 = 15 \times 3$ $\newline$ $y - 5 = 45$ $\newline$ Now, add $5$ to both sides to solve for y. $\newline$ $y - 5 + 5 = 45 + 5$ $\newline$ $y = 50$
Substitute y into equation: Now that we have the value of y, we can substitute it into the first equation to solve for x. $\newline$ Given: $y = (x - 5)^2 + 1$ $\newline$ Substitute y with $50$ . $\newline$ $50 = (x - 5)^2 + 1$ $\newline$ Now, subtract $1$ from both sides to isolate the squared term. $\newline$ $50 - 1 = (x - 5)^2 + 1 - 1$ $\newline$ $49 = (x - 5)^2$ $\newline$ Next, take the square root of both sides to solve for x. $\newline$ $\sqrt{49} = \sqrt{(x - 5)^2}$ $\newline$ $x - 5 = \pm7$ $\newline$ This gives us two possible solutions for x when we add $5$ to both sides. $\newline$ $x = 5 + 7$ or $x = 5 - 7$ $\newline$ $x = 12$ or $x = -2$

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