Question 499336
{{{f(x)=x^2 -5x-6}}}  or {{{y=x^2 -5x-6}}}


The general form of a quadratic is "{{{y = ax^2 + bx + c}}}". 

For graphing, the leading coefficient "{{{a}}}" indicates how "{{{fat}}}" or how "{{{skinny}}}" the {{{parabola}}} will be.

For |a| > {{{1}}} (such as {{{a = 3}}} or {{{a = -4}}}), the parabola will be "{{{skinny}}}", because it {{{grows}}} more {{{quickly}}} (three times as fast or four times as fast, respectively, in the case of our sample values of a). 

For | a | < {{{1}}} (such as {{{a = 1/3}}} or {{{a = -1/4}}}), the parabola will be "{{{fat}}}", because it {{{grows}}} more {{{slowly}}} (one-third as fast or one-fourth as fast, respectively, in the examples). 

Also, if {{{a}}} is {{{negative}}}, then the parabola is {{{upside-down}}}.

In your case,{{{a}}} is {{{positive}}} and the parabola is {{{upward}}}.

If the quadratic is written in the form {{{y=a(x-h)^2+k}}} (the vertex form), then the vertex is the point ({{{h}}}, {{{k}}}). 

For a given quadratic {{{y = ax^2 + bx + c}}}, the vertex ({{{h}}}, {{{k}}}) is found by computing {{{h=-b/2a}}} , and {{{k=(4ac-b^2)/4a}}}.

In your case, {{{y=x^2 -5x-6}}},  the vertex ({{{h}}}, {{{k}}}) will be:

{{{h=-b/2a}}}

{{{h=-(-5)/2*1}}}

{{{h =5/2}}}

{{{h =2.5}}}

{{{k=(4ac-b^2)/4a}}}

{{{k=(4*1*(-6)-(-5)^2)/4*1}}}

{{{k=(-24-25)/4}}}

{{{k =-49 / 4}}}

{{{k =-12.25}}}

so, 

the vertex ({{{h}}},{{{ k}}}) is at ({{{2.5}}}, {{{-12.25}}})


the "{{{axis}}} of {{{symmetry}}}"

If you look at a parabola, you'll notice that you could {{{draw}}} a {{{vertical}}} line right up through the middle which would split the parabola into two mirrored halves. This vertical line, right {{{through}}} the {{{vertex}}}, is called the {{{axis}}} of {{{symmetry}}}. 

If you're asked for the axis, write down the line "{{{x = h}}}", where {{{h}}} is just the {{{x-coordinate}}} of the vertex. So in the example above, then the axis would be the vertical line {{{x = h = 2.5}}}.


{{{ graph( 500, 500, -10, 10, -15, 10, x^2 -5x-6) }}}