Answer
No real zeros.
No real $x$-intercepts.
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 $H(x)=0$:
$$4x^2+x+1=0$$
Comparing $4x^2+x+1$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$
$$\therefore a = 4, b=1 , c =1$$
Evaluating the discriminant $b^2-4ac$
$$b^2-4ac = (1)^2-4 \times 4 \times 1 = -15$$
Since the discriminant is negative, then the function has no real zeros.
