Question

what is the maximum y value of the graph of y = -3x^2 + 2x - 5

Answer 1

To find the maximum \( y \) value of the quadratic function \( y = -3x^2 + 2x - 5 \), we can use the vertex formula for a parabola. For a quadratic in the form \( y = ax^2 + bx + c \), the x-coordinate of the vertex (which gives the maximum or minimum value depending on the sign of \( a \)) is given by:

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

In this function, \( a = -3 \) and \( b = 2 \).

Plugging in the values:

\[
x = -\frac{2}{2(-3)} = -\frac{2}{-6} = \frac{1}{3}
\]

Now, we need to find the maximum \( y \) value by substituting \( x = \frac{1}{3} \) back into the equation:

\[
y = -3\left(\frac{1}{3}\right)^2 + 2\left(\frac{1}{3}\right) - 5
\]

Calculating \( y \):

\[
y = -3\left(\frac{1}{9}\right) + \frac{2}{3} - 5
\]
\[
y = -\frac{3}{9} + \frac{2}{3} - 5
\]
\[
y = -\frac{1}{3} + \frac{2}{3} - 5
\]
\[
y = \frac{1}{3} - 5
\]
\[
y = \frac{1 - 15}{3} = \frac{-14}{3}
\]

So, the maximum \( y \) value of the graph of \( y = -3x^2 + 2x - 5 \) is \( \frac{-14}{3} \).