Question 171206
{{{f(x)=x^2-4x-5 }}} ---> {{{ax^2+bx+c=0}}}
Okay, first we go for the Vertex-form of quadratic: {{{f(x)=a(x-h)^2+k}}}
where, h & k are {{{x,y}}} points on the vertex.
We'll do this by completing the square: We take {{{1/2}}} of "b" in the quadratic eq'n. and squared it. In this case it, "b"=4, so half of it --> {{{(1/2)(4)=2}}}, then squared, {{{2^2=highlight(4)}}}.
So we need to add "4" to the eqn:
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{{{f(x)=(x^2-4x+4)highlight(-4)-5}}} ----> remember to subtract 4 too so the expression does not change.
{{{f(x)=(x-2)^2-9}}} ----> follows our vertex form now, where {{{h=2}}} & {{{k=-9}}}. We'll see in the graph:
{{{drawing(300,300,-10,10,-11,10,grid(1),graph(300,300,-10,10,-11,10,x^2-4x-5),blue(circle(2,-9,.20))))}}} ----> see points (2,-9). This will be the vertex.
Now we'll see the x-intercepts by solving the eqn:
{{{x^2-4x-5=0}}}
By Quadratic Formula, where ---->{{{system(a=1,b=-4,c=-5)}}}
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}
{{{x=(-(-4)+-sqrt(-4^2-4*1*-5))/(2*1)}}}
{{{x=(4+-sqrt(16+20))/2=(4+-sqrt(36))/2=(4+-6)/2}}}
{{{x=(4+6)/2=10/2=highlight(5)}}}
{{{x=(4-6)/2=-2/2=highlight(-1)}}}
X-intercepts---->{{{system(x=5,x=-1)}}}
See the graph again:
{{{drawing(300,300,-5,7,-11,10,grid(1),graph(300,300,-5,7,-11,10,x^2-4x-5),blue(circle(2,-9,.20)),green(circle(5,0,.20)),green(circle(-1,0,.20)))}}}
.
Thank you,
Jojo