Answer
Zeros: $\dfrac{-1+\sqrt{17}}{2}, \dfrac{-1-\sqrt{17}}{2}$
$x$-intercepts: $\dfrac{-1+\sqrt{17}}{2}, \dfrac{-1-\sqrt{17}}{2}$
Work Step by Step
To find the zeros of a function $f$, solve the equation $f(x)=0$
The zeros of the function are also the $x-$intercepts.
Let $f(x)=0$:
$$x^2+x-4=0$$
Comparing $x^2+x-4=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$
$$\therefore a =1, b=1, c =-4$$
Evaluating the discriminant $b^2-4ac$
$$b^2-4ac = (-1)^2-4 \times 1\times -4 = 17$$
The quadratic formula is given by:
$$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x= \dfrac{-1\pm \sqrt{17}}{2\times 1}$$
$$x=\dfrac{-1\pm \sqrt{17}}{2}$$
Thus, the zeros, which are also the $x$-intercepts, are $\dfrac{-1+\sqrt{17}}{2}$ and $\dfrac{-1-\sqrt{17}}{2}$.
