a. Find the derivative function $$ f^{\prime} $$ for the function $$ f $$. b. Find an equation of the line tangent to the graph of $$ f $$ at $$ (a, f(a)) $$ for the given value of $$ f(x)=2 x^{2}-x-3, a=2 $$ a. $$ f^{\prime}(x)=\square $$

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a. $$ f^{\prime}(x) = 4x - 1 $$ b. The equation of the tangent line is $$ y = 7x - 11 $$. To find the derivative function $$ f' $$ for the function $$ f(x) = 2x^2 - x - 3 $$, we will differentiate $$ f(x) $$ with respect to $$ x $$. ### a. Finding the derivative function $$ f' $$ Using the power rule for differentiation, we have: 1. The derivative of $$ 2x^2 $$ is $$ 4x $$. 2. The derivative of $$ -x $$ is $$ -1 $$. 3. The derivative of the constant $$ -3 $$ is $$ 0 $$. Putting it all together, we get: $$ f'(x) = 4x - 1 $$ ### b. Finding the equation of the tangent line at $$ (a, f(a)) $$ where $$ a = 2 $$ First, we need to evaluate $$ f(2) $$: $$ f(2) = 2(2^2) - 2 - 3 = 2(4) - 2 - 3 = 8 - 2 - 3 = 3 $$ So the point on the graph is $$ (2, 3) $$. Next, we need to find $$ f'(2) $$: $$ f'(2) = 4(2) - 1 = 8 - 1 = 7 $$ Now we have the slope of the tangent line, which is $$ 7 $$, and the point $$ (2, 3) $$. We can use the point-slope form of the equation of a line: $$ y - y_1 = m(x - x_1) $$ Substituting $$ m = 7 $$, $$ x_1 = 2 $$, and $$ y_1 = 3 $$: $$ y - 3 = 7(x - 2) $$ Now, we can simplify this equation: $$ y - 3 = 7x - 14 $$ $$ y = 7x - 11 $$ ### Summary a. The derivative function is: $$ f'(x) = 4x - 1 $$ b. The equation of the tangent line at $$ (2, 3) $$ is: $$ y = 7x - 11 $$

a. Find the derivative function \( f^{\prime} \) for the function \( f \). b. Find an equation of the line tangent to the graph of \( f \) at \( (a, f(a)) \) for the given value of \( f(x)=2 x^{2}-x-3, a=2 \) a. \( f^{\prime}(x)=\square \)

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a. Find the derivative function \( f^{\prime} \) for the function \( f \). b. Find an equation of the line tangent to the graph of \( f \) at \( (a, f(a)) \) for the given value of \( f(x)=2 x^{2}-x-3, a=2 \) a. \( f^{\prime}(x)=\square \)

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