Question

a parabola opening up or down has vertex (0, 1) and passes through (12, -11). write its equation in vertex form. simplify any fractions.

Explanation:

Step1: Recall vertex form of parabola

The vertex form of a parabola that opens up or down is \( y = a(x - h)^2 + k \), where \((h,k)\) is the vertex. Given the vertex \((0,1)\), so \( h = 0 \) and \( k = 1 \). Substituting these values into the vertex form, we get \( y = a(x - 0)^2 + 1 \), which simplifies to \( y = ax^2 + 1 \).

Step2: Substitute the point \((12, -11)\)

We know the parabola passes through the point \((12, -11)\). Substitute \( x = 12 \) and \( y = -11 \) into the equation \( y = ax^2 + 1 \):

$$ -11 = a(12)^2 + 1 $$

Step3: Solve for \( a \)

First, calculate \( 12^2 = 144 \). So the equation becomes:

$$ -11 = 144a + 1 $$

Subtract 1 from both sides:

$$ -11 - 1 = 144a $$

$$ -12 = 144a $$

Now, divide both sides by 144 to solve for \( a \):

$$ a = \frac{-12}{144} = -\frac{1}{12} $$

Step4: Write the final equation

Substitute \( a = -\frac{1}{12} \), \( h = 0 \), and \( k = 1 \) back into the vertex form \( y = a(x - h)^2 + k \):

$$ y = -\frac{1}{12}(x - 0)^2 + 1 $$

Simplifying, we get:

$$ y = -\frac{1}{12}x^2 + 1 $$

Answer:

\( y = -\frac{1}{12}x^2 + 1 \)