This is the graph, on the x and y axis od the graph you plot the point at 4 (positive)
Is the equation like this? x^2 + 25y^2?
If the equation is like that, the answer would be (x-5y)^2 or (x-5y)(x-5y).
how did I get this? Just divide the factors or the get the square root of 25y^2 and it would be 5y. Then just try if it is positive or negative. In this case, it is negative because there is no factor w/ no exponential form. I hopemy answer helped you.
1) y= - 2x² + 8x. It's a parabola open downward (a<0)
2) x - 2.23.y + 10.34 = 0 . Re-write it : y = (x/2.23) + (10.34/2.23), a linear equation.
To find the intersections between 1) & 2), let 1) = 2)
-2x² + 8x = (x/2.23) + (10.34/2.23)
-2x² + 8x - (x/2.23) - (10.34/2.23) =0 ; solve this quadratic for x values:
x' (that is A) = 0.772 & x" (that is B) = 3. (these are the values of x-intercept (parabola with line). To calculate the y-values, plug x' & x' in the equation:
for x' = 0.772, y = 0.34 → B(0.772 , 0.34)
for x" = 3, y = 0.016 → A(3 , 0.O16)
So B IS AT 0.34 Unit from the ground
x 1 x 1 = 