A parabola representing a quadratic equation has x-intercepts at (1,0)

Question

(
1
,
0
)
and (−3,0)
(
−
3
,
0
)
and a third point with coordinates at (−1,4)
(
−
1
,
4
)
. Which quadratic equation in intercept form accurately represents this information?(1 point)
Responses

y=1(x−1)(x+3)
𝑦
=
1
(
𝑥
−
1
)
(
𝑥
+
3
)
y is equal to 1 open paren x minus 1 close paren open paren x plus 3 close paren

y=−1(x+1)(x+3)
𝑦
=
−
1
(
𝑥
+
1
)
(
𝑥
+
3
)
y is equal to negative 1 open paren x plus 1 close paren open paren x plus 3 close paren

y=−1(x−1)(x+3)
𝑦
=
−
1
(
𝑥
−
1
)
(
𝑥
+
3
)
y is equal to negative 1 open paren x minus 1 close paren open paren x plus 3 close paren

y=−1(x−1)(x−3)

Answer 1

To find the quadratic equation in intercept form representing the parabola with x-intercepts at \((1,0)\) and \((-3,0)\), we can start from the intercept form of a quadratic equation:

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

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

For this problem:

• \(r_1 = 1\)
• \(r_2 = -3\)

So the equation becomes:

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

Next, we can find the value of \(a\) using the third point \((-1, 4)\). We plug the x-value and y-value of this point into the equation to solve for \(a\):

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

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

2. Calculate the terms inside the parentheses:

\[ 4 = a(-2)(2) \] \[ 4 = -4a \]

3. Solve for \(a\):

\[ a = -1 \]

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

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

Thus, the quadratic equation in intercept form is:

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

This matches the third option in your list.