Let's analyze and solve the two mathematical problems shown in the image.

1. Koordinat titik balik kurva

The given quadratic function is:

$f(x) = x^2 - 10x + 29$

To find the vertex (titik balik) of the parabola, we use the formula for the x-coordinate of the vertex:

$x_{\text{vertex}} = -\frac{b}{2a}$

Here, $a = 1$ and $b = -10$. Substituting into the formula:

$x_{\text{vertex}} = -\frac{-10}{2(1)} = 5$

Now, substitute $x = 5$ into the function to find the y-coordinate of the vertex:

$f(5) = (5)^2 - 10(5) + 29 = 25 - 50 + 29 = 4$

Thus, the vertex (titik balik) is at the point $(5, 4)$.

Answer: The correct option is $(5, 4)$.

2. Perhatikan grafik fungsi kuadrat berikut

The second question asks to identify the quadratic function that matches the given graph.

The graph shows a parabola that opens upwards with these points:

• $(-3, 0)$, $(-1, -4)$, $(1, 0)$, and $(0, -3)$.

The general form of a quadratic function is:

$f(x) = ax^2 + bx + c$

From the graph, the parabola crosses the x-axis at $(-3, 0)$ and $(1, 0)$. This means the factors of the quadratic function are $(x + 3)(x - 1)$, giving us the equation:

$f(x) = a(x + 3)(x - 1)$

Expanding this:

$f(x) = a(x^2 - x - 3x - 3) = a(x^2 - 2x - 3)$

Now, to determine the value of $a$, we use the point $(-1, -4)$. Substituting $x = -1$ and $f(x) = -4$:

$-4 = a((-1)^2 - 2(-1) - 3) = a(1 + 2 - 3) = a(0)$

Since this equation doesn't provide the value of $a$, I would recommend trying this again with all points