Find the maximum $$ y $$-value on the graph of $$ y=f(x) $$. $$ f(x)=-4 x^{2}+x-7 $$ The maximum $$ y $$-value is $$ y=\square $$. (Type an integer or a fraction.)

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To find the maximum $$ y $$-value of the quadratic function $$ f(x) = -4x^2 + x - 7 $$, we can use the vertex formula for a parabola given by $$ y = ax^2 + bx + c $$. The vertex $$ x $$-coordinate can be found using the formula:

$$
x = -\frac{b}{2a}
$$

In our function, we have:
- $$ a = -4 $$
- $$ b = 1 $$
- $$ c = -7 $$

Substituting the values of $$ a $$ and $$ b $$ into the vertex formula:

$$
x = -\frac{1}{2 \cdot -4} = -\frac{1}{-8} = \frac{1}{8}
$$

Next, we substitute $$ x = \frac{1}{8} $$ back into the function $$ f(x) $$ to find the maximum $$ y $$-value:

$$
f\left(\frac{1}{8}\right) = -4\left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right) - 7
$$

Calculating $$ \left(\frac{1}{8}\right)^2 $$:

$$
\left(\frac{1}{8}\right)^2 = \frac{1}{64}
$$

Now substituting this back into the function:

$$
f\left(\frac{1}{8}\right) = -4 \cdot \frac{1}{64} + \frac{1}{8} - 7
$$

Calculating $$ -4 \cdot \frac{1}{64} $$:

$$
-4 \cdot \frac{1}{64} = -\frac{4}{64} = -\frac{1}{16}
$$

Now substituting this value into the equation:

$$
f\left(\frac{1}{8}\right) = -\frac{1}{16} + \frac{1}{8} - 7
$$

Next, we convert $$ \frac{1}{8} $$ to sixteenths:

$$
\frac{1}{8} = \frac{2}{16}
$$

Now we can combine the terms:

$$
f\left(\frac{1}{8}\right) = -\frac{1}{16} + \frac{2}{16} - 7 = \frac{1}{16} - 7
$$

To combine $$ \frac{1}{16} $$ and $$ -7 $$, we convert $$ -7 $$ to sixteenths:

$$
-7 = -\frac{112}{16}
$$

Now we can combine:

$$
f\left(\frac{1}{8}\right) = \frac{1}{16} - \frac{112}{16} = \frac{1 - 112}{16} = \frac{-111}{16}
$$

Thus, the maximum $$ y $$-value is:

$$
\boxed{-\frac{111}{16}}
$$
The maximum $$ y $$-value is $$ y=-\frac{111}{16} $$.

Find the maximum -value on the graph of .

The maximum -value is . (Type an integer or a fraction.)

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