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 (4,3)
(
4
,
3
)
. 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 minus 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 minus 3 close paren

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)

Answer 1

To determine the correct quadratic equation in intercept form, we start with the intercept form of a quadratic equation, which is:

\[ y = a(x - p)(x - q) \]

where \( p \) and \( q \) are the x-intercepts.

Given the x-intercepts at (1, 0) and (3, 0), we have:

• \( p = 1 \)
• \( q = 3 \)

Thus, the equation in intercept form so far is:

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

Now, we also have a third point (4, 3). We can use this point to find the value of \( a \).

Substituting \( x = 4 \) and \( y = 3 \) into the equation, we get:

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

This simplifies to:

\[ 3 = a(3)(1) \]

So,

\[ 3 = 3a \] \[ a = 1 \]

Thus, the equation becomes:

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

The correct option is:

\[ y = 1(x - 1)(x - 3) \] or simply \( y = (x - 1)(x - 3) \).

Answer 2

A parabola representing a quadratic equation has x-intercepts at (1,0)
(
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 negative 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 1 open paren x minus 1 close paren open paren x plus 3 close paren

y=−1(x−1)(x−3)