Question #9a872

1 Answer
Jan 20, 2018

The graphs of the two equations intersect at the points (6,23) and (3,14)

Explanation:

The formatting seems to have taken a hit there, so we'll assume that the equations are

y=x^2-6x+23 and
y-5=3x.

The second equation can be written as y=3x+5. Both equations need to be true for specific values of x and y, so we can substitute the value of y in terms of x in the first equation to solve for x:

3x+5=x^2-6x+23=>x^2-9x+18=0

This is just a quadratic equation, so let's find the discriminant:

D=(-9)^2-4*1*18=81-72=9>0

Since the discriminant is greater than 0, we know the equation has two distinct roots.

x_1=(9+3)/2=6

x_2=(9-3)/2=3

Then we just need to use those values of x to determine y using y=3x+5. If x=6, y=23 and if x=3, y=14

Therefore, the graphs of the two equations intersect at the points (6,23) and (3,14)

NOTE: For any equation of the form ax^2+bx+c=0, the discriminant is calculated by D=b^2-4ac and the roots are calculated by (-b+-sqrtD)/(2a) if D>=0, and if D<0 there are no real roots.