Question

Question #9a872

Answer 1

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.