Asked by julian celeste
A parabola representing a quadratic equation has x-intercepts at (1,0)
and (3,0)
and a third point with coordinates at (4,3)
. Which quadratic equation in intercept form accurately represents this information?(1 point)
Responses
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 is equal to 1 open paren x plus 1 close paren open paren x minus 3 close paren
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)
and (3,0)
and a third point with coordinates at (4,3)
. Which quadratic equation in intercept form accurately represents this information?(1 point)
Responses
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 is equal to 1 open paren x plus 1 close paren open paren x minus 3 close paren
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)
Answers
Answered by
GPT-4o mini
To find the quadratic equation in intercept form that matches the given x-intercepts (1, 0) and (3, 0) and passes through the point (4, 3), we start with the intercept form of a quadratic equation:
\[ y = a(x - p)(x - q) \]
where \( p \) and \( q \) are the x-intercepts. In this case:
- \( p = 1 \)
- \( q = 3 \)
Thus, the equation in intercept form becomes:
\[ y = a(x - 1)(x - 3) \]
Next, we need to determine the value of \( a \) using the point (4, 3). Substituting \( x = 4 \) and \( y = 3 \) into the equation gives:
\[ 3 = a(4 - 1)(4 - 3) \]
Calculating the values inside the parentheses:
\[ 3 = a(3)(1) \]
\[ 3 = 3a \]
\[ a = 1 \]
Now we substitute \( a \) back into the equation:
\[ y = 1(x - 1)(x - 3) \]
So the quadratic equation that represents the information is:
\[ \boxed{y = 1(x - 1)(x - 3)} \]
\[ y = a(x - p)(x - q) \]
where \( p \) and \( q \) are the x-intercepts. In this case:
- \( p = 1 \)
- \( q = 3 \)
Thus, the equation in intercept form becomes:
\[ y = a(x - 1)(x - 3) \]
Next, we need to determine the value of \( a \) using the point (4, 3). Substituting \( x = 4 \) and \( y = 3 \) into the equation gives:
\[ 3 = a(4 - 1)(4 - 3) \]
Calculating the values inside the parentheses:
\[ 3 = a(3)(1) \]
\[ 3 = 3a \]
\[ a = 1 \]
Now we substitute \( a \) back into the equation:
\[ y = 1(x - 1)(x - 3) \]
So the quadratic equation that represents the information is:
\[ \boxed{y = 1(x - 1)(x - 3)} \]
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