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 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 indeed on the graph of $f$.

To find the derivative, we differentiate $f(x)$:

$f'(x) = 3x^2 + 2x - 2.$
Next, we find the slope of the tangent at $x = 1$: $f'(1) = 3(1)^2 + 2(1) - 2 = 3 + 2 - 2 = 3.$
Thus, the slope is $m = 3$.

Using the point-slope form of the equation of a line, we have:

$y - y_1 = m(x - x_1)$ where $(x_1, y_1) = (1, 7)$ and $m = 3$.

Substituting these values gives:

$y - 7 = 3(x - 1).$
Simplifying this:

$y - 7 = 3x - 3 \\y = 3x + 4.$
Thus, the equation of the tangent line is $y = 3x + 4$.