For f(x)=3x^2+9x+8 find the equation of the tangent line at the following values of x. (a) x=1 The equation of the tangent line at x=1 is square (b) x=-1 The equation of the tangent line at x=-1 is square

To find the equation of the tangent line to the function $f(x)=3x^{2}+9x+8$ at a given value of $x$, we need to find the derivative of the function and then evaluate it at the given value of $x$ to find the slope of the tangent line. Finally, we can use the point-slope form of a linear equation to find the equation of the tangent line.<br /><br />Let's start with part (a) where $x=1$.<br /><br />Step 1: Find the derivative of the function $f(x)$.<br />The derivative of $f(x)$ is given by:<br />$f'(x) = \frac{d}{dx}(3x^{2}+9x+8) = 6x+9$<br /><br />Step 2: Evaluate the derivative at $x=1$ to find the slope of the tangent line.<br />$f'(1) = 6(1)+9 = 15$<br /><br />Step 3: Use the point-slope form of a linear equation to find the equation of the tangent line.<br />The point-slope form of a linear equation is given by:<br />$y-y_1 = m(x-x_1)$<br />where $m$ is the slope of the line and $(x_1, y_1)$ is a point on the line.<br /><br />In this case, the slope of the tangent line is $15$, and we can choose the point $(1, f(1))$ on the line. So, the equation of the tangent line at $x=1$ is:<br />$y-f(1) = 15(x-1)$<br /><br />Step 4: Substitute the value of $f(1)$ into the equation.<br />$f(1) = 3(1)^{2}+9(1)+8 = 20$<br />So, the equation of the tangent line at $x=1$ is:<br />$y-20 = 15(x-1)$<br /><br />Simplifying the equation, we get:<br />$y = 15x + 5$<br /><br />Therefore, the equation of the tangent line at $x=1$ is $y = 15x + 5$.<br /><br />Now let's move on to part (b) where $x=-1$.<br /><br />Step 1: Find the derivative of the function $f(x)$.<br />The derivative of $f(x)$ is given by:<br />$f'(x) = \frac{d}{dx}(3x^{2}+9x+8) = 6x+9$<br /><br />Step 2: Evaluate the derivative at $x=-1$ to find the slope of the tangent line.<br />$f'(-1) = 6(-1)+9 = 3$<br /><br />Step 3: Use the point-slope form of a linear equation to find the equation of the tangent line.<br />The point-slope form of a linear equation is given by:<br />$y-y_1 = m(x-x_1)$<br />where $m$ is the slope of the line and $(x_1, y_1)$ is a point on the line.<br /><br />In this case, the slope of the tangent line is $3$, and we can choose the point $(-1, f(-1))$ on the line. So, the equation of the tangent line at $x=-1$ is:<br />$y-f(-1) = 3(x+1)$<br /><br />Step 4: Substitute the value of $f(-1)$ into the equation.<br />$f(-1) = 3(-1)^{2}+9(-1)+8 = 2$<br />So, the equation of the tangent line at $x=-1$ is:<br />$y-2 = 3(x+1)$<br /><br />Simplifying the equation, we get:<br />$y = 3x + 5$<br /><br />Therefore, the equation of the tangent line at $x=-1$ is $y = 3x + 5$.