How do you find the x and y intercepts for y = -2(x-1)^2 + 3y=2(x1)2+3?

2 Answers
Apr 3, 2018

yy intercept is 1
xx intercepts are 2.22 and -0.22

Explanation:

The yy intercept is when x =0x=0
=> y = -2(0 -1)^2 +3y=2(01)2+3

=> y = -2 y=2xx× 1 +3#

=> y = -2 +3y=2+3

=> y = 1y=1

The xx intercept is when y = 0y=0
=> 0 = -2(x -1)^2 +30=2(x1)2+3

=> 0 = -2 (x^2 -2x +1) +30=2(x22x+1)+3

=> 0 = -2x^2 +4x -2 +30=2x2+4x2+3

=> 2x^2 -4x -1 = 02x24x1=0

Use the formula x=(-b\pm sqrt(b^2 -4ac))/(2a)x=b±b24ac2a

x=(4\pm sqrt(16+8))/(4)x=4±16+84

x=(4 +sqrt(24))/4x=4+244 or x = (4 - sqrt(24))/4x=4244
x = 2.22474487x=2.22474487 or x = -0.22474487x=0.22474487

Apr 3, 2018

p_(x_1)(1+sqrt(3/2)|0) or p_(x_2)(1-sqrt(3/2)|0)px1(1+320)orpx2(1320)
p_y(0|1)py(01)

Explanation:

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

y = -2(x-1)^2 + 3y=2(x1)2+3
x=0x=0
y = -2(0-1)^2 + 3y=2(01)2+3
y = -2 + 3=1y=2+3=1
p_y(0|1)py(01)

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

0 = -2(x-1)^2 + 3|-30=2(x1)2+33
-3 = -2(x-1)^2 |:(-2)3=2(x1)2:(2)
3/2=(x-1)^2|sqrt()32=(x1)2
+-sqrt(3/2)=x-1|+1±32=x1+1
1+-sqrt(3/2)=x1±32=x
p_1(1+sqrt(3/2)|0) or p_2(1-sqrt(3/2)|0)p1(1+320)orp2(1320)

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