The function $f : x ightarrow x^3 + x^2 - 2x + 7$ is defined for $x \\in \\mathbb{R}$ - Leaving Cert Mathematics - Question 4 - 2016

We need to substitute $x = 1$ into the function:

$f(1) = 1^3 + 1^2 - 2(1) + 7 = 1 + 1 - 2 + 7 = 7.$

Since $f(1) = 7$, the point $A(1, 7)$ is on the graph of $f$.

To find the derivative:

$f'(x) = 3x^2 + 2x - 2.$

Now, substituting $x = 1$ to find the slope:

$f'(1) = 3(1)^2 + 2(1) - 2 = 3 + 2 - 2 = 3.$

Thus, the slope of the tangent at $x = 1$ is $3$.

Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, where $m$ is the slope we found and $(x_1, y_1) = (1, 7)$:

$y - 7 = 3(x - 1).$

This can be rearranged to:

$y = 3x + 4$

or equivalently, $3x - y + 4 = 0.$