Question 258148
First, we'll graph the function: {{{7-6x-x^2 = 0}}}
{{{graph(400,400,-10,5,-5,20,-x^2-6x+7)}}}
a) The "solutions" for a quadratic equation mean...for what values of the independent variable (x) is the function (y) equal to zero.
By inspecting the graph, you can easily see that the curve (a parabola) intersects the x-axis (where y = 0) at x = 1 and x = -7, so these are the solutions.
b) The function does have a "maximum" as you can see on the graph where the direction of the curve changes from "up" to "down" or the slope of the curve is zero.  Another way to determine if a quadratic function, such as you have here, has a maximum or minimum is to examine the coefficient of the {{{x^2}}} term:
If positive, the parabola opens upward and the function has a "minimum".
If negative (as is the case here) the parabola opens downward and the function has a "maximum".
The "axis (line) of symmetry" is the vertical line that passes through the vertex and the vertex is the point (maximum or minimum) at which the slope is zero. The equation of this vertical line is: {{{highlight(x = -3)}}}
From the inspection of the graph of this function, you can see that the vertex (maximum) is located at the point (-3, 16)
You can also find the location of the x-coordinate of the vertex algbraically from:
{{{x = (-b)/2a}}} Substitute a = -1 and b = -6.
{{{x = (-(-6))/2(-1)}}}
{{{highlight(x = -3)}}} and, by substituting this into the given quadratic equation, you can determine the y-coordinate.
{{{y = 7-6x-x^2}}} Substitute x = -3.
{{{y = 7-6(-3)-(-3)^2}}}
{{{y = 7+18-9}}}
{{{y = 16}}}
The coordinates of the vertex are: (-3, 16)