Question 202316
find (a)the directrix, (b)the focus, and (c)the roots of the parabola 
{{{y = x^2 - 5x + 4}}}
<pre><font size = 4 color = "indigo"><b>

We have to get it in the form 

{{{(x-h)^2=4p(y-k)}}}


{{{y = x^2 - 5x + 4}}}

Swap sides:

{{{x^2 - 5x + 4 = y}}}

Add -4 to both sides the get the x-terms alone on the left.

{{{x^2 - 5x=y-4}}}

Multiply the coefficient of x, which is {{{-5}}} by {{{1/2}}}
This gives {{{-5/2}}}.  Now we square {{{-5/2}}} and get {{{""+25/4}}}
We add {{{""+25/4}}} to both sides:

{{{x^2 - 5x+25/4=y-4+25/4}}}

Factor the left side, and write the {{{-4}}} as {{{-16/4}}}

{{{(x-5/2)(x-5/2)=y-16/4+25/4}}}

Write the left side as the square of a binomial,
combine the fractions on the right:

{{{(x-5/2)^2=y+9/4}}}

Now so that equation will look like this:

{{{(x-h)^2=4p(y-k)}}}

we put the right side in parentheses and put a 1
coefficient before the parentheses, like this:

{{{(x-5/2)^2=1(y+9/4)}}}

Now we can compare it to the equation:

{{{(x-h)^2=4p(y-k)}}}

and get {{{-h=-5/2}}}, so {{{h=5/2}}}

{{{-k=""+9/4}}}, so {{{k=-9/4}}}

So the vertex is ({{{h}}},{{{k}}}) or ({{{5/2}}},{{{-9/4}}})

And {{{4p=1}}}, so {{{p=1/4}}}

Now let's begin by plotting the vertex, which is ({{{5/2}}},{{{3/2}}}),

But for plotting purposes, mixed numbers are better
than improper fractions, so for plotting vertex ({{{5/2}}},{{{-9/4}}}),
we rewrite it as ({{{2&1/2}}},{{{-2&1/4}}})


{{{drawing(400,400,-1,6,-3,4,
graph(400,400,-1,6,-3,4), 
line(5/2-.05,-9/4,5/2+.05,-9/4), line(5/2,-9/4-.05,5/2,-9/4+.05),
locate(5/2,-9/4,"(5/2,-9/4)")
         )}}}

Now we will find the x-intercepts, by settng {{{y=0}}}
in the original equation, and finding the "roots":

{{{y = x^2 - 5x + 4}}}
{{{0 = x^2 - 5x + 4}}}
{{{0 = (x-1)(x-4)}}}
{{{x-1=0}}}, so {{{x=1}}}
{{{x-4=0}}}, so {{{x=4}}}

So the x-intecepts are (1,0) and (4,0) 

So we plot those:

{{{drawing(400,400,-1,6,-3,4,
graph(400,400,-1,6,-3,4), 
line(5/2-.05,-9/4,5/2+.05,-9/4), line(5/2,-9/4-.05,5/2,-9/4+.05),
locate(5/2,-9/4,"(5/2,-9/4)"),
line(1-.05,0,1+.05,0), line(1,0-.05,1,0+.05),
line(4-.05,0,4+.05,0), line(4,0-.05,4,0+.05)

         )}}}

and sketch in the parabola:

{{{drawing(400,400,-1,6,-3,4,
graph(400,400,-1,6,-3,4,x^2 - 5x + 4), 
line(5/2-.05,-9/4,5/2+.05,-9/4), line(5/2,-9/4-.05,5/2,-9/4+.05),
locate(5/2,-9/4,"(5/2,-9/4)"),
line(1-.05,0,1+.05,0), line(1,0-.05,1,0+.05),
line(4-.05,0,4+.05,0), line(4,0-.05,4,0+.05)

         )}}}

Now the focus is p units from the vertex INSIDE
the parabola, so since the parabola opens upward,
we add P or {{{1/4}}} to the y-coordinate of the
vertex. Since the vertx is ({{{5/2}}},{{{-9/4}}}),
the focus = ({{{5/2}}},{{{-9/4+1/4}}}) = ({{{5/2}}},{{{-8/4}}}) 
= ({{{5/2}}},{{{-2}}})

So we draw that point:

{{{drawing(400,400,-1,6,-3,4,
graph(400,400,-1,6,-3,4,x^2 - 5x + 4), 
line(5/2-.05,-9/4,5/2+.05,-9/4), line(5/2,-9/4-.05,5/2,-9/4+.05),
locate(5/2-.5,-2+.4,"(5/2,-2)"),
line(1-.05,0,1+.05,0), line(1,0-.05,1,0+.05),
line(4-.05,0,4+.05,0), line(4,0-.05,4,0+.05),

line(5/2-.05,-2,5/2+.05,-2), line(5/2,-2-.05,5/2,-2+.05)

         )}}}

Now the directrix is a line OUTSIDE the parabola which is
also p-units, or {{{1/4}}} from the vertex.

Since the y-coordinate of the vertex is {{{-9/4}}} we want
the directrix to be {{{1/4}}} unit below the vertex, so we
subtract {{{-9/4-1/4=-10/4=-5/2}}}

So the directrix is the horizontal line whose equation is 

{{{y=-5/2}}}.  I'll draw it in in green:

{{{drawing(400,400,-1,6,-3,4,
graph(400,400,-1,6,-3,4,x^2 - 5x + 4,-5/2), 
line(5/2-.05,-9/4,5/2+.05,-9/4), line(5/2,-9/4-.05,5/2,-9/4+.05),
locate(5/2-.5,-2+.4,"(5/2,-2)"),
line(1-.05,0,1+.05,0), line(1,0-.05,1,0+.05),
line(4-.05,0,4+.05,0), line(4,0-.05,4,0+.05),
locate(5/2,-2.7,"y=-5/2"),
line(5/2-.05,-2,5/2+.05,-2), line(5/2,-2-.05,5/2,-2+.05)

         )}}}

So the focus is the POINT ({{{5/2}}},{{{-2}}}) and the 

directrix is the LINE {{{y=-5/2}}}

The "roots" are really the y-coordinates of the 
x-intercepts or 1 and 4.

Edwin</pre>