Question 178329
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1)We have to remember when 2 lines are perpendicular, the {{{slope=m[2]}}} of the other one is "negative reciprocal" of the 1st :------>{{{m2=-1/m[1]}}}
Arranging Line {{{y-2x+5}}} via Slope-intercept form, {{{y=mx+b}}}:
{{{y=2x+5}}}
We can see it has a {{{Slope=m[1]=2}}}, then the line that passes thru point (-1,3) has a slope:
{{{m[2]=-1/2}}}
Via Slope-intercept Form: point (-1,3)
{{{3=(-1/2)(-1)+b}}}
{{{3=(1/2)+b}}}---->{{{b=3-(1/2)=(6-1)/2=5/2}}}, Y-Intercept
Therefore, line eqn. {{{y=(-1/2)x+(5/2)}}} ----> {{{y+(1/2)x=5/2}}} (Standard form) passes thru point (-1,3) and perpendicular to line {{{y-2x=5}}}
we'll see:
{{{drawing(400,400,-5,5,-5,8,grid(1),graph(400,400,-5,5,-5,8,2x+5, (-1/2)x+(5/2)),blue(circle(-1,3,.15)))}}}-----> RED Line>>>> {{{y=2x+5}}}; GREEN Line>>>> {{{y+(1/2)x=5/2}}}
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2)Sketch a graph for a quadratic function given as below:
{{{f (x) = x² - x - 6}}}
Solving the quadratic:
{{{x^2-x-6=0}}}
where{{{system(a=1,b=-1,c-6)}}}
{{{x=(-b+-sqrt(b^2-4ac))/2a=(-(-1)+-sqrt(-1^2-4*1*-6))/(2*1)}}}
{{{x=(1+-sqrt(1+24))/2=(1+-sqrt(25))/2}}}
{{{x=(1+sqrt(25))/2=(1+5)/2=6/2=highlight(3)}}}
{{{x=(1-sqrt(25))/2=(1-5)/2=-4/2=highlight(-2)}}}
X-intercepts--------> (3,0) & (-2,0)
{{{drawing(400,400,-5,5,-8,5,grid(1),graph(400,400,-5,5,-8,5,x^2-x-6), blue(circle(3,0,.12)),blue(circle(-2,0,.12)))}}}

Thank you,
Jojo</pre>