The question in the image asks for the value of $k$ such that the graph of the quadratic function $f(x) = kx^2 + kx + 1$ touches the x-axis.

For the graph of a quadratic function to touch the x-axis (i.e., to have exactly one solution or root), the discriminant of the quadratic equation must be zero.

The general form of a quadratic equation is:

$ax^2 + bx + c = 0$

where the discriminant $\Delta$ is:

$\Delta = b^2 - 4ac$

If $\Delta = 0$, the quadratic equation has one root, meaning the graph touches the x-axis.

In this case, the quadratic equation is:

$f(x) = kx^2 + kx + 1$

So, $a = k$, $b = k$, and $c = 1$.

Now, the discriminant $\Delta$ becomes:

$\Delta = k^2 - 4(k)(1)$

Simplifying this:

$\Delta = k^2 - 4k$

For the graph to touch the x-axis, we set $\Delta = 0$:

$k^2 - 4k = 0$

Factoring the equation:

$k(k - 4) = 0$

This gives two solutions:

$k = 0 \quad \text{or} \quad k = 4$

Thus, the values of $k$ for which the graph touches the x-axis are $k = 0$ or $k = 4$.

The correct answer is E. 0 atau 4.

Do you want further explanation or have any questions?

Here are 5 related questions for practice:

1. What is the discriminant and how does it determine the number of roots?
2. What happens to the graph if the discriminant is positive or negative?
3. How do you find the roots of a quadratic function when the discriminant is non-zero?
4. Can the value of $k$ change the direction of the parabola?
5. How would you factor the quadratic expression if $k = 4$?

Tip: When solving quadratic equations, always consider the discriminant to understand how many roots the equation will have.