Question
Which of the following is a point you can generate to create the parable for the quarterback of equation y = -6x^2 + 3x + 8
Answers
GPT-4o mini
To find the vertex of the quadratic equation \( y = -6x^2 + 3x + 8 \), we can use the vertex formula for a parabola given by \( y = ax^2 + bx + c \). The x-coordinate of the vertex is found using the formula:
\[
x = -\frac{b}{2a}
\]
In your equation, \( a = -6 \) and \( b = 3 \).
Calculating the x-coordinate of the vertex:
\[
x = -\frac{3}{2(-6)} = -\frac{3}{-12} = \frac{1}{4}
\]
Now, to find the y-coordinate of the vertex, we substitute \( x = \frac{1}{4} \) back into the equation:
\[
y = -6\left(\frac{1}{4}\right)^2 + 3\left(\frac{1}{4}\right) + 8
\]
Calculating \( \left(\frac{1}{4}\right)^2 \):
\[
\left(\frac{1}{4}\right)^2 = \frac{1}{16}
\]
Substituting that into the equation:
\[
y = -6\left(\frac{1}{16}\right) + \frac{3}{4} + 8
\]
Calculating \( -6 \times \frac{1}{16} \):
\[
-6 \times \frac{1}{16} = -\frac{6}{16} = -\frac{3}{8}
\]
So the equation becomes:
\[
y = -\frac{3}{8} + \frac{3}{4} + 8
\]
Converting \( \frac{3}{4} \) to eighths to combine:
\[
\frac{3}{4} = \frac{6}{8}
\]
So now:
\[
y = -\frac{3}{8} + \frac{6}{8} + 8 = \frac{3}{8} + 8
\]
To convert 8 to eighths:
\[
8 = \frac{64}{8}
\]
Thus:
\[
y = \frac{3}{8} + \frac{64}{8} = \frac{67}{8}
\]
Now we have the vertex:
\[
\left( \frac{1}{4}, \frac{67}{8} \right)
\]
This point is the vertex of the parabola and can be used to sketch the graph of the quadratic function. The parabola opens downward since the coefficient of \( x^2 \) (i.e., \( a = -6 \)) is negative.
\[
x = -\frac{b}{2a}
\]
In your equation, \( a = -6 \) and \( b = 3 \).
Calculating the x-coordinate of the vertex:
\[
x = -\frac{3}{2(-6)} = -\frac{3}{-12} = \frac{1}{4}
\]
Now, to find the y-coordinate of the vertex, we substitute \( x = \frac{1}{4} \) back into the equation:
\[
y = -6\left(\frac{1}{4}\right)^2 + 3\left(\frac{1}{4}\right) + 8
\]
Calculating \( \left(\frac{1}{4}\right)^2 \):
\[
\left(\frac{1}{4}\right)^2 = \frac{1}{16}
\]
Substituting that into the equation:
\[
y = -6\left(\frac{1}{16}\right) + \frac{3}{4} + 8
\]
Calculating \( -6 \times \frac{1}{16} \):
\[
-6 \times \frac{1}{16} = -\frac{6}{16} = -\frac{3}{8}
\]
So the equation becomes:
\[
y = -\frac{3}{8} + \frac{3}{4} + 8
\]
Converting \( \frac{3}{4} \) to eighths to combine:
\[
\frac{3}{4} = \frac{6}{8}
\]
So now:
\[
y = -\frac{3}{8} + \frac{6}{8} + 8 = \frac{3}{8} + 8
\]
To convert 8 to eighths:
\[
8 = \frac{64}{8}
\]
Thus:
\[
y = \frac{3}{8} + \frac{64}{8} = \frac{67}{8}
\]
Now we have the vertex:
\[
\left( \frac{1}{4}, \frac{67}{8} \right)
\]
This point is the vertex of the parabola and can be used to sketch the graph of the quadratic function. The parabola opens downward since the coefficient of \( x^2 \) (i.e., \( a = -6 \)) is negative.