Question

y = a(1-2)^2 + 1

Explanation:

⚡ Using what you learned: graphing quadratic functions

Step 1: Identify the vertex of the parabola

From the given graph, the lowest point (vertex) of the parabola is located at:

$$ (h, k) = (1, 0) $$

Step 2: Use the vertex form of a quadratic equation

The vertex form is:

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

Substituting the vertex \( (1, 0) \):

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

$$ y = a(x - 1)^2 $$

Step 3: Find the value of \( a \) using another point

Identify another clear point on the graph, such as the y-intercept:

$$ (x, y) = (0, 1) $$

Substitute \( x = 0 \) and \( y = 1 \) into the equation to solve for \( a \):

$$ 1 = a(0 - 1)^2 $$

$$ 1 = a(-1)^2 $$

$$ 1 = a(1) $$

$$ a = 1 $$

Step 4: Write the final equation

Substitute \( a = 1 \) back into the equation:

$$ y = (x - 1)^2 $$

Answer:

$$ (x - 1)^2 $$