Find the area of the intersection coordinates a, B and △ AOB of the straight line y = 2x + 8 and the parabola y = x & # 178

Find the area of the intersection coordinates a, B and △ AOB of the straight line y = 2x + 8 and the parabola y = x & # 178

Make height from a and B to X axis respectively, intersect at C and D
y=2x+8
y=x²
The solution is x = - 2,4
∴ A(-2,4),B(4,16)
The area of triangle AOB can be expressed as s trapezoid abcd-s △ aod-s △ BOD
S trapezoid ABCD = (4 + 16) × (2 + 4) × (1 / 2) = 60
S△AOB=60-2×4×(1／2)-4×16×(1／2)=24

Find the intersection coordinates of parabola y = x & # 178; - 2x-5 and straight line y = x + 5

Solution;
y=x²-2x-5
y=x+5
x²-2x-5=x+5
x²-3x-10=0
(x-5)(x+2)=0
X = 5 or x = - 2
So the intersection is (5,10) or (- 2,3)

The intersection of parabola y = - x2-2x + 3 and X axis is___ The point of intersection with y axis is___ ．

∵ when x = 0, y = 3, ∵ and y-axis intersection is (0, 3); ∵ when y = 0, - x2-2x + 3 = 0, the solution is: x = - 3 or 1, ∵ parabolic y = - x2-2x + 3 and X-axis intersection is (- 3, 0), (1, 0); so the answer is (- 3, 0), (1, 0), (0, 3)

Find the coordinates of the intersection of parabola y = x & # 178; + 7x + 3 and straight line y = 2x + 9

X & # 178; + 7x + 3 = 2x + 9, we get X1 = - 6, X2 = 1, and substitute it into any curve equation to get the intersection coordinates (- 6, - 3) (1,11)