# Find the intercepts of y= 2x^2 -4x+8.

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### 2 Answers

We have the equation. First let's find the x - intercept. Here y = 0

0 = 2x^2 - 4x+8

=> x^2 - 2x + 8 = 0

=> x^2 - 4x + 2x - 8 = 0

=> x(x - 4) + 2(x - 4) = 0

=> (x + 2)(x - 4) = 0

The x-intercepts are (-2 , 0) and (4 , 0)

For the y-intercept, x = 0

=> y = 2x^2 - 4x+8

=> y = 8

The y - intercept is ( 0, 8)

**The intercepts of the graph are (0 , 8) , (-2 , 0) and (4 , 0)**

Given the curve y= 2x^2 - 4x +8

We need to find the x and y intercepts.

First we will determine the y-intercepts.

The y-intercept is when the curve meets the y-axis, then the value of x is 0.

==> y= 0+ 8 = 8

Then the y-intercept is (0, 8).

Now we will determine the x-intercept.

Then y= 0.

==> 2x^2 -4x +8 = 0

==> (2x+4)(x-4) = 0

==> x= -2

==> x= 4

Then we have two x-intercepts ( -2, 0) and (4, 0)

**Then the intercepts are:**

**y-intercept (0, 8) **

**x-intercepts (-2, 0) and (4,0)**