Find the points where the parabola y=x²+4x+3 intersects with the straight line y=-x+9
To find the points where the parabola y=x²+4x+3 intersects with the straight line y=-x+9, we can set the two equations equal to each other and solve for x.
Setting y = -x+9 equal to the parabola equation y = x²+4x+3:
-x+9 = x²+4x+3
Combine like terms:
x²+5x-6 = 0
Factor the quadratic equation:
(x+6)(x-1) = 0
Setting each factor equal to zero:
x+6 = 0 or x-1 = 0
Solving for x:
x = -6 or x = 1
Now that we have the values of x, we can substitute these values back into either equation to find the corresponding y-values.
For x = -6:
y = (-6)²+4(-6)+3
y = 36-24+3
y = 15
The point of intersection for x = -6 is (-6, 15).
For x = 1:
y = (1)²+4(1)+3
y = 1+4+3
y = 8
The point of intersection for x = 1 is (1, 8).
Therefore, the parabola y=x²+4x+3 intersects with the straight line y=-x+9 at the points (-6, 15) and (1, 8).
To find the points of intersection between the parabola y=x²+4x+3 and the straight line y=-x+9, we need to solve the equations y=x²+4x+3 and y=-x+9 simultaneously.
Substituting the value of y from the equation y=-x+9 into the equation y=x²+4x+3, we get:
-x+9 = x²+4x+3
Rearranging the equation, we have:
x²+5x-6 = 0
We can factorize this quadratic equation, like so:
(x+6)(x-1) = 0
Setting each factor equal to zero, we have:
x+6=0 or x-1=0
Solving these equations, we find:
x=-6 or x=1
Now, substituting these values of x into the equation y=-x+9, we can find the corresponding y-values.
For x=-6:
y=-(-6)+9 = 6+9 = 15
So, one point of intersection is (-6,15).
And for x=1:
y=-(1)+9 = 8
So, the other point of intersection is (1,8).
Therefore, the parabola y=x²+4x+3 intersects the straight line y=-x+9 at the points (-6,15) and (1,8).