Question 163607
Congratulations for making an attempt at solving this problem!
Given:
{{{y = -x^2+6x-5}}}
Find:
a) The vertex of the parabola:
The x-coordinate is given by:
{{{x = -b/2a}}} In your equation, a = -1 and b = 6, so...
{{{x = -6/2(-1)}}}
{{{x = 3}}} You got this part correct! Now substitute this into the original equation to find the y-coordinate.
{{{y = -(3)^2+6(3)-5}}}
{{{y = -9+18-5}}}
{{{y = 4}}} You also got this correct!
The vertex is located at (3, 4) but please note, in your calculations, you show...{{{(-3)^2 = -9}}} and this is not correct.
Remember that: {{{(-3)^2 = (-3)(-3)}}}={{{9}}} but, {{{-(3)^2 = -(3)(3)}}}={{{-9}}}
The axis of symmetry is the vertical line that bisects the parabola and, of course, it runs right through the vertex, so its equation would be:
{{{x = 3}}}
The maximum or minimum value?
Well, notice that the coefficient of the {{{x^2}}} term is negative, so this indicates that the parabola opens downward, hence, the vertex will be a maximum, and its location is, as you have already found, (3, 4).
Let's see what the graph looks like for this equation:
{{{graph(400,400,-2,8,-5,5,-x^2+6x-5)}}}