How do you find the x and y intercepts for y = -2(x-1)^2 + 3?

2 Answers
Apr 3, 2018

y intercept is 1
x intercepts are 2.22 and -0.22

Explanation:

The y intercept is when x =0
=> y = -2(0 -1)^2 +3

=> y = -2 xx 1 +3#

=> y = -2 +3

=> y = 1

The x intercept is when y = 0
=> 0 = -2(x -1)^2 +3

=> 0 = -2 (x^2 -2x +1) +3

=> 0 = -2x^2 +4x -2 +3

=> 2x^2 -4x -1 = 0

Use the formula x=(-b\pm sqrt(b^2 -4ac))/(2a)

x=(4\pm sqrt(16+8))/(4)

x=(4 +sqrt(24))/4 or x = (4 - sqrt(24))/4
x = 2.22474487 or x = -0.22474487

Apr 3, 2018

p_(x_1)(1+sqrt(3/2)|0) or p_(x_2)(1-sqrt(3/2)|0)
p_y(0|1)

Explanation:

Lets start with the interception of the y-axis. At the point where the graph intercepts this axis, x=0. Furthermore, it is the only point where x=0.

y = -2(x-1)^2 + 3
x=0
y = -2(0-1)^2 + 3
y = -2 + 3=1
p_y(0|1)

It works the same way for the x-axis. The only difference is that y=0 instead of x=0.

0 = -2(x-1)^2 + 3|-3
-3 = -2(x-1)^2 |:(-2)
3/2=(x-1)^2|sqrt()
+-sqrt(3/2)=x-1|+1
1+-sqrt(3/2)=x
p_1(1+sqrt(3/2)|0) or p_2(1-sqrt(3/2)|0)

graph{y=-2(x-1)^2+3 [-2, 4, -5, 5]}