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What are the x-intercepts of #x²-y+2x=6 #?

Answer 1

The x-intercepts are at #( -1+sqrt7,0)# and #( -1-sqrt7,0)#

We first isolate y.

#x^2-y+2x=6# #-y+2x=6-x^2# #-y=6-x^2-2x# #y=-6+x^2+2x# #y=x^2+2x-6#

We now try to find the zeros. Since we cannot factor this equation in any quick way, we find the two zeros using the quadratic formula #(-b+-sqrt(b^2-4(a)(c)))/(2(a))# #a=1# #b=2# #c=-6#

The zeros are: #(-2+-sqrt(2^2-4(1)(-6)))/(2(1))# #(-2+-sqrt(4+24))/2# #(-2+-sqrt28)/2# #(-2+-2sqrt7)/2# #-1+-sqrt7#

#-1+-sqrt7# are the x values for the x-intercepts.

Note that we know that(and it will, if you are correct) when we plug in the zeros, we get zero for the equation.

*Here is the idea of how the zeros are the answer. Since the y value is 0, we know that the point must lie on the x coordinate, which means those points are the x intercepts.

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Answer 2

Eva Abramson

To find the x-intercepts, set y to 0 and solve for x.

x² - y + 2x = 6
x² - 0 + 2x = 6
x² + 2x = 6
x² + 2x - 6 = 0
(x + 3)(x - 2) = 0

x = -3, 2

The x-intercepts are -3 and 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.