Question 414468
{{{y=-x^2+5x-4}}}....factor

{{{y=-x^2+x+4x-4}}}

{{{y=(-x^2+x)+(4x-4)}}}

{{{y=-x(x-1)+4(x-1)}}}

{{{y=(x-1)(4-x)}}}


This is a quadratic equation (degree of 2), so

First, find x-intercepts, 

Let{{{ y = 0}}} and solve for {{{x}}}:

{{{(x-1)(4-x)=0}}}

{{{x-1=0}}}........->....{{{x = 1}}} and 

{{{4-x=0}}}........->....{{{x = 4}}}

So, the {{{x-intercepts}}} (roots) are (4,0) and (1,0)


Now, find the {{{y-intercepts}}}, let {{{x = 0}}} and solve for {{{y}}},

{{{(0-1)(4-0)=-4}}}, so the {{{y-intercept}}} is (0,-4)


The axis of {{{symmetry}}} is just the average of the two {{{x-intercepts}}},

{{{(4 + 1)/2= 2.5}}}

So, the axis of symmetry is {{{x = 2.5}}}

Since the sign of {{{x^2}}} is negative then this function will have a maximum.

Now, to find it's maximum (vertex).

Formula for maximum value is {{{x = -b/2a}}}

{{{a = -1}}}, {{{b = 5}}}, {{{c = -4}}} ( Coefficients of {{{y = -x^2 + 5x- 4 }}})

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

this is exactly {{{2.5}}}, which is not surprising because the {{{maximum}}}{{{ occurs}}} on the axis of {{{symmetry}}}.

Now, find the {{{y}}} value corresponding to the {{{x}}} value,

Substitute, {{{x = 2.5}}} into the equation and solve for {{{y}}},

{{{y=(2.5-1)(4-2.5)}}}

{{{y = (1.5)(1.5)}}}

{{{y = 2.25}}}


Summarize:

You should have more than enough now for a graph,

{{{x-intercepts}}}: (4,0) and (1,0)

{{{y-intercept}}}: (0,-4)

maximum value (vertex) : (2.5,2.25)

axis of {{{symmetry}}}: {{{x = 2.5}}}


{{{ graph( 500, 500, -10, 10, -10, 5, -x^2+5x-4) }}}