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Question

Straight line #y-x=7# intersects with #y-x ^ 2= -1# curve at points A and B. Find the coordinates of points A and B

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Answer 1 · 2018-07-10T13:42:39

#((1-sqrt33)/2,4.63)# and #((1+sqrt33)/2,10.37)# are your coordinates A and B

This question is asking for the points of intersection or at which points do your line and curve meet.

#y-x=7#
#y=x+7# --- (1)

#y-x^2=-1#
#y=x^2-1# --- (2)

If you have points of intersection, that means your curve and line must meet and have the same point.

Therefore, you can let the equation of your line and curve equal to one another

(1)=(2)
#x+7=x^2-1#
#x^2-x-8=0#

Using the quadratic formula,
#x=(1+-sqrt(1+32))/2#

#x=(1+-sqrt33)/2#

#x=(1-sqrt33)/2# or #x=(1+sqrt33)/2#

When #x=(1-sqrt33)/2#, #y=4.63# ---> #((1-sqrt33)/2,4.63)#
When #x=(1+sqrt33)/2#, #y=10.37# ---> #((1+sqrt33)/2,10.37)#

Below is the graph
graph{(y-x-7)(y-x^2+1)=0 [-10, 10, -5, 5]}

Answer 2 · 2018-07-10T13:47:30

#x_1=1/2*(1-sqrt(33)),y_1=1/2*(15-sqrt(33))#
#x_2=1/2(1+sqrt(33)),y_2=1/2*(15+sqrt(33))#

From the second equation we get

#y=x+7# substituting this in the seond equation:
#x+7-x^2=-1#
so we get

#x^2-x-8=0#
solving this equation by the quadratic formula:

#x_(1,2)=1/2pm\sqrt{1/4+8)#

#1/4+8=(1+32)/4=33/4#

so we get

#x_(1,2)=1/2pm sqrt(33)/2#
and we get

#y_1=7+x_1#
#y_2=7+x_2#