Respuesta :
the intersection points is where both equations' graph meet, or namely where one equation equals the other, or for this case when y = y, so
[tex]\begin{cases} y = 4x - 5\\ y = 2x^2-5x-10 \end{cases}\qquad \stackrel{y}{4x-5}~~ = ~~\stackrel{y}{2x^2-5x-10} \\\\\\ -5=2x^2-9x-10\implies 0=2x^2-9x-5 \\\\\\ 0=(x-5)(2x+1)\implies \begin{cases} 0 = x-5\\ 5 = x\\[-0.5em] \hrulefill\\ 0 = 2x+1\\ -1=2x\\ -\frac{1}{2}=x \end{cases}[/tex]
well, we know what "x" is at those point, to get the "y" value, we can use either of the equations and substitute our "x" in it, hmmmm let's use the 1st equation for that
[tex]y = 4(\stackrel{x}{5})-5\implies y = 20-5\implies y = 15 \\\\\\ y = 4\stackrel{x}{\left( -\frac{1}{2} \right)}-5\implies y = -2-5\implies y = -7 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{intersection points}}{(5~~,~~15)\qquad \left(-\frac{1}{2}~~,~-7 \right)}~\hfill[/tex]
The intersection points of the given system would be at the point:
[tex](5, 15) (-1/2, -7)[/tex]
The two equations,
[tex]y=4x-5[/tex] [tex]...(i)[/tex]
[tex]y=2x^2-5x-10[/tex] [tex]...(ii)[/tex]
so,
[tex]-5 = 2x^2 - 9x - 10[/tex] ⇒ [tex]0 = 2x^2 - 9x - 5[/tex]
we get,
[tex](x - 5) (2x + 1) = 0[/tex]
∵ [tex]x = 5 and x = -1/2[/tex]
Now solving for [tex]y[/tex] by putting the value of [tex]x[/tex] in equation (i),
[tex]y = 4(5) - 5[/tex]
∵ [tex]y = 15[/tex]
[tex]y = 4(-1/2)^2 - 5[/tex]
∵ [tex]y = -7[/tex]
Thus, the values are [tex](5, 15) (-1/2, -7)[/tex]
Learn more about "Intersection Point" here:
brainly.com/question/13373561
