Question 321729
The general equation for a parabola is, {{{y=ax^2+bx+c}}}
Plugging in the points you get a system of equations,
1.{{{a-b+c=0}}}
2.{{{4a+2b+c=0}}}
3.{{{c=-16}}}
Substitute eq. 3 into eq. 1 and eq. 2,
4.{{{a-b=16}}}
5.{{{4a+2b=16}}}
Multiply eq. 4 by 2 and add to eq. 5 to solve for a.
{{{2a-2b+4a+2b=32+16}}}
{{{6a=48}}}
{{{a=8}}}
Then from eq. 4,
{{{8-b=16}}}
{{{b=-8}}}
The parabola is then,
{{{y=8x^2-8x-16}}}
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For the line, calculate the slope,
{{{m=(y2-y1)/(x2-x1)=(30-(-10))/(4-(-5))}}}
{{{m=40/9}}}
Using the point slope form of a line, {{{y=mx+b}}}, use either point to solve for b.
{{{y=(40/9)x+b}}}
{{{30=(40/9)(4)+b}}}
{{{b=270/9-160/9}}}
{{{b=110/9}}}
The line is then,
{{{y= (40/9)x+110/9}}}
To find the intersection points set the two function equal to each other and solve for x.
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{{{drawing(300,300,-20,20,-20,40,grid(1),circle(-5,-10,.5),circle(4,30,0.5),circle(-1,0,.5),circle(2,0,.5),circle(0,-16,0.5),graph(300,300,-20,20,-20,40, (40/9)x+110/9, 8x^2-8x-16))}}}
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{{{8x^2-8x-16=(40/9)x+110/9}}}
{{{72x^2-72x-144=40x+110}}}
{{{72x^2-112x-254=0}}}
Use the quadratic formula,
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (112 +- sqrt(112^2-4*72*(-254) ))/(2*72) }}}
{{{x = (112 +- sqrt(12544+73152 ))/(144) }}}
{{{x = (112 +- sqrt(85696 ))/(144) }}}
{{{x = (112 +- 8sqrt(1339 ))/(144) }}}
{{{x = (14 +- sqrt(1339 ))/(18) }}}
Now that you have the x values, use either equation to solve for y.