Question
In the xy-coordinate plane, the graph of the equations y = 3x^2 −12x −36 has zeros at x = and x =, where d<e The graph has a minimum at (__, -48).
Answers
GPT-4o mini
To find the zeros of the quadratic equation \( y = 3x^2 - 12x - 36 \), we can set the equation equal to zero and solve for \( x \):
\[
3x^2 - 12x - 36 = 0
\]
First, we can simplify the equation by dividing all terms by 3:
\[
x^2 - 4x - 12 = 0
\]
Next, we can factor this equation. We are looking for two numbers that multiply to \(-12\) and add up to \(-4\). The numbers \(-6\) and \(2\) fit this requirement:
\[
(x - 6)(x + 2) = 0
\]
Setting each factor to zero gives us the zeros of the equation:
\[
x - 6 = 0 \quad \Rightarrow \quad x = 6
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]
Thus, the zeros of the function \( y = 3x^2 - 12x - 36 \) are \( x = -2 \) and \( x = 6 \), where \( d = -2 \) and \( e = 6 \).
Next, we find the vertex of the parabola, which provides the minimum point because the coefficient of \( x^2 \) is positive (indicating the parabola opens upwards). The x-coordinate of the vertex (\( x_{\text{vertex}} \)) for a quadratic equation in the form \( ax^2 + bx + c \) is given by:
\[
x_{\text{vertex}} = -\frac{b}{2a}
\]
In our case, \( a = 3 \) and \( b = -12 \):
\[
x_{\text{vertex}} = -\frac{-12}{2 \times 3} = \frac{12}{6} = 2
\]
Now that we have the x-coordinate of the vertex, we can find the corresponding y-coordinate by substituting \( x = 2 \) back into the original equation:
\[
y = 3(2^2) - 12(2) - 36
\]
\[
y = 3 \times 4 - 24 - 36
\]
\[
y = 12 - 24 - 36
\]
\[
y = 12 - 60 = -48
\]
Therefore, the minimum point of the graph is at \( (2, -48) \).
Summarizing everything, we have:
- The zeros are \( x = -2 \) and \( x = 6 \) (where \( d < e \)).
- The minimum occurs at \( (2, -48) \).
Thus, the final answer is:
The zeros are \( x = -2 \) and \( x = 6 \), and the graph has a minimum at \( (2, -48) \).
\[
3x^2 - 12x - 36 = 0
\]
First, we can simplify the equation by dividing all terms by 3:
\[
x^2 - 4x - 12 = 0
\]
Next, we can factor this equation. We are looking for two numbers that multiply to \(-12\) and add up to \(-4\). The numbers \(-6\) and \(2\) fit this requirement:
\[
(x - 6)(x + 2) = 0
\]
Setting each factor to zero gives us the zeros of the equation:
\[
x - 6 = 0 \quad \Rightarrow \quad x = 6
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]
Thus, the zeros of the function \( y = 3x^2 - 12x - 36 \) are \( x = -2 \) and \( x = 6 \), where \( d = -2 \) and \( e = 6 \).
Next, we find the vertex of the parabola, which provides the minimum point because the coefficient of \( x^2 \) is positive (indicating the parabola opens upwards). The x-coordinate of the vertex (\( x_{\text{vertex}} \)) for a quadratic equation in the form \( ax^2 + bx + c \) is given by:
\[
x_{\text{vertex}} = -\frac{b}{2a}
\]
In our case, \( a = 3 \) and \( b = -12 \):
\[
x_{\text{vertex}} = -\frac{-12}{2 \times 3} = \frac{12}{6} = 2
\]
Now that we have the x-coordinate of the vertex, we can find the corresponding y-coordinate by substituting \( x = 2 \) back into the original equation:
\[
y = 3(2^2) - 12(2) - 36
\]
\[
y = 3 \times 4 - 24 - 36
\]
\[
y = 12 - 24 - 36
\]
\[
y = 12 - 60 = -48
\]
Therefore, the minimum point of the graph is at \( (2, -48) \).
Summarizing everything, we have:
- The zeros are \( x = -2 \) and \( x = 6 \) (where \( d < e \)).
- The minimum occurs at \( (2, -48) \).
Thus, the final answer is:
The zeros are \( x = -2 \) and \( x = 6 \), and the graph has a minimum at \( (2, -48) \).