Question

A parable representing a quarterback equation has an x-intercept that one, 0 and -3, 0 in a third point with coordinates at negative 1, 4 with quadratic equation intercept form actually represents this information y = -1 x - 1 x + 3 y = 1 X - 1 x + 3

Answer 1

It seems like you're trying to construct a quadratic equation in intercept form based on the given x-intercepts and another point. Let's break down the information you've provided.

### Given Information

1. **X-Intercepts**: (1, 0) and (-3, 0)
2. **Third Point**: (-1, 4)

### Quadratic Form

The intercept form of a quadratic equation can be expressed as:

\[ y = a(x - r_1)(x - r_2) \]

where \( r_1 \) and \( r_2 \) are the x-intercepts.

### Step 1: Set Up the Equation

Using the given x-intercepts (1 and -3), the equation in intercept form is:

\[ y = a(x - 1)(x + 3) \]

### Step 2: Expand the Equation

We can expand this:

\[ y = a[(x - 1)(x + 3)] \]

Using the distributive property:

\[ y = a[x^2 + 3x - x - 3] \]
\[ y = a[x^2 + 2x - 3] \]

### Step 3: Use the Third Point to Solve for \( a \)

Now, we will use the third point (-1, 4) to find the value of \( a \).

Substitute \( x = -1 \) and \( y = 4 \):

\[ 4 = a[(-1)^2 + 2(-1) - 3] \]
\[ 4 = a[1 - 2 - 3] \]
\[ 4 = a[-4] \]

Now, solve for \( a \):

\[ a = \frac{4}{-4} \]
\[ a = -1 \]

### Final Equation

Now we substitute \( a \) back into our equation:

\[ y = -1(x - 1)(x + 3) \]

Expanding this gives:

\[ y = -1(x^2 + 2x - 3) \]
\[ y = -x^2 - 2x + 3 \]

### Summary

The quadratic equation that fits the given x-intercepts and the point (-1, 4) is:

\[ y = -x^2 - 2x + 3 \]

If you have any further queries or need clarification on any point, feel free to ask!