Question

You are asked to graph a parabola given the equation y=14x2−2x+7 . At which point would you graph the vertex of the parabola?

Answer 1

To find the vertex of the parabola given by the equation \( y = 14x^2 - 2x + 7 \), we can use the vertex formula for a quadratic equation in standard form \( y = ax^2 + bx + c \). The x-coordinate of the vertex can be found using the formula:

\[
x = -\frac{b}{2a}
\]

In this case, \( a = 14 \) and \( b = -2 \). Plugging in these values:

\[
x = -\frac{-2}{2 \cdot 14} = \frac{2}{28} = \frac{1}{14}
\]

Next, we need to find the y-coordinate of the vertex by substituting \( x = \frac{1}{14} \) back into the equation:

\[
y = 14\left(\frac{1}{14}\right)^2 - 2\left(\frac{1}{14}\right) + 7
\]

Calculating each term:

1. \( 14\left(\frac{1}{14}\right)^2 = 14 \cdot \frac{1}{196} = \frac{14}{196} = \frac{1}{14} \)
2. \( -2\left(\frac{1}{14}\right) = -\frac{2}{14} = -\frac{1}{7} \)
3. \( 7 = \frac{98}{14} \) (converting 7 to have a common denominator of 14)

Now, adding these values together:

\[
y = \frac{1}{14} - \frac{1}{7} + \frac{98}{14}
\]

The term \( -\frac{1}{7} \) can be expressed with a denominator of 14:

\[
-\frac{1}{7} = -\frac{2}{14}
\]

Now substituting:

\[
y = \frac{1}{14} - \frac{2}{14} + \frac{98}{14} = \frac{1 - 2 + 98}{14} = \frac{97}{14}
\]

Thus, the vertex of the parabola is located at:

\[
\left( \frac{1}{14}, \frac{97}{14} \right)
\]

So, you would graph the vertex of the parabola at the point \( \left( \frac{1}{14}, \frac{97}{14} \right) \).