We are given the quadratic function $f(x) = ax^2 + bx + c$ and the following information:

• The vertex is at $(0, 1)$, meaning $f(0) = 1$.
• The graph passes through the point $(1, 9)$, meaning $f(1) = 9$.

Step 1: Use the vertex information

Since the vertex is $(0, 1)$, we can immediately determine $c$ because at $x = 0$, $f(0) = 1$. So, using $f(0) = ax^2 + bx + c$:

$f(0) = a(0)^2 + b(0) + c = c = 1$

Thus, $c = 1$.

Step 2: Use the point $(1, 9)$

Next, we use the fact that the point $(1, 9)$ lies on the curve, so $f(1) = 9$. Substituting $x = 1$ and $f(1) = 9$ into the equation:

$f(1) = a(1)^2 + b(1) + 1 = 9$ $a + b + 1 = 9$ $a + b = 8 \quad \text{(Equation 1)}$

Step 3: Use the vertex form condition

For a parabola with vertex $(h, k)$, the equation is given by:

$f(x) = a(x - h)^2 + k$

In this case, the vertex is $(0, 1)$, so:

$f(x) = a(x - 0)^2 + 1 = ax^2 + 1$

Thus, from the vertex form, we know $b = 0$.

Step 4: Solve for $a$

Now that we know $b = 0$, we substitute it into Equation 1:

$a + 0 = 8 \implies a = 8$

Final Answer:

Thus, the values of $a$, $b$, and $c$ are: $a = 8, \quad b = 0, \quad c = 1$

The function is $f(x) = 8x^2 + 1$.

Would you like further clarification, or should I dive deeper into any part of this?

---

Here are 5 related questions for practice:

1. What is the axis of symmetry of the function $f(x) = 8x^2 + 1$?
2. How do you find the vertex of a quadratic function in standard form?
3. If the function passed through $(2, 33)$ instead of $(1, 9)$, how would that affect the equation?
4. Can a quadratic function have more than one vertex?
5. What is the domain and range of the function $f(x) = 8x^2 + 1$?

Tip: The vertex form of a quadratic function is useful for finding the vertex directly from the equation.