How would you find the intercept for the equation y^2=2-3x-2x^2? I need the steps to understand
1 answer:
9514 1404 393
Answer:
- x-intercepts: -2, +1/2
- y-intercepts: ±√2
Step-by-step explanation:
You always find the intercepts by setting the known coordinate to its known value and solving for the value of the unknown coordinate.
An x-intercept is always found on the line y=0. Setting y=0 and solving for x, we find ...
0 = 2 -3x -2x^2
x^2 +3/2x -1 = 0 . . . . . . . . . . . . . . . . divide by -2
(x2^2 +3/2x +(3/4)^2) -1 -(3/4)^2 = 0 . . . . . complete the square
(x +3/4)^2 = 25/16 . . . . . add 25/16 and write as a square
x +3/4 = ±5/4 . . . . . . . take the square root
x = -3/4 ±5/4 = -8/4, +2/4
The x-intercepts are -2 and +1/2.
__
A y-intercept is always found on the line x=0. Setting x=0 and solving for y, we find ...
y^2 = 2 -0 -0
y = ±√2
The y-intercepts are -√2 and +√2.
_____
A graphing calculator can give you a good idea of what the intercepts are.
You might be interested in
Answer:
(x + 1)² + (y + 3)² = 16
Step-by-step explanation:
The equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the centre and r is the radius
Given
x² + y² + 2x + 6y - 6 = 0
Collect the x and y terms together and add 6 to both sides
x² + 2x + y² + 6y = 6
To complete the square
add ( half the coefficient of the x/ y terms )² to both sides
x² + 2(1)x + 1 + y² + 2(3)y + 9 = 6 + 1 + 9
(x + 1)² + (y + 3)² = 16
with centre = (- 1, - 3) and r = = 4