Answer: (2,2) and (-5,16)
Step-by-step explanation:
Here we have both Line (Linear Function) and Parabola (Quadratic Function)
So I am gonna write these equations here,
[tex]y=x^2+x-4\\y=-2x+6[/tex]
The first equation has Parabola graph (Since it's second degree.)
and the second equation has line graph.
To find the intersection, you have to substitute either -2x+6 in first equation (Quadratic) or x^2+x-4 in second equation (Linear)
For me, I am going to substitute x^2+x-4 in y=-2x+6.
[tex]x^2+x-4=-2x+6[/tex]
Now solve the equation and find the value of x.
Since it's Quadratic Equation (Because there's x^2) I'd move -2x+6 to the left side.
[tex]x^2+x-4+2x-6=0[/tex] Finish things here (Subtract and Addition)
[tex]x^2+3x-10=0[/tex] What two numbers multiply to 10? Find the factors of 10, that are [1 and 10] and [2 and 5]
Now think about it, do you think that if 1 and 10 subtract or even addition, do you think that it'd be 3? No, of course not.
So 2 and 5 is right.
[tex](x-2)(x+5)=0[/tex] (5-2 = 3) and (5*(-2) = -10)
Then we get both x, [tex]x=2,-5[/tex]
However, this is not it. You have to substitute both x in Linear Equation.
Substitute x = 2 in y=-2x+6
[tex]y=-2(2)+6\\y=-4+6\\y=2[/tex]
Order = [tex](2,2)[/tex]
Then substitute x = -5 in y=-2x+6
[tex]y=-2(-5)+6\\y=10+6\\y=16[/tex]
Order = [tex](-5,16)[/tex]
So the intersections are both (2,2) and (-5,16) as shown in graph below.
