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What are the x-intercept(s) of #y =-x^2-2x+5#?

Answer 1

Julia Adams

x-intercepts: #x=sqrt(6)-1# and #x=-sqrt(6)-1#

The x-intercepts are the values of #x# when #y=0# (the line of the graph crosses the X-axis when #y=0#)

#y=-x^2-2x+5=0#

#rArrx^2+2x-5 = 0#

Using the quadratic formula #color(white)("XXX")x=(-2+-sqrt(2^2-4(1)(-5)))/(2(1))#

#color(white)("XXXX")= (-2+-sqrt(24))/2#

#color(white)("XXXX")=(-2+-2sqrt(6))/2#

#color(white)("XXXX")=-1+-sqrt(6)#

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Answer 2

Kennedy Adams

To find the x-intercepts of (y = -x^2 - 2x + 5), set (y) to 0 and solve for (x).

[0 = -x^2 - 2x + 5]

This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:

[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}]

For (y = -x^2 - 2x + 5), (a = -1), (b = -2), and (c = 5). Plug these values into the formula:

[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-1)(5)}}{2(-1)}] [x = \frac{2 \pm \sqrt{4 + 20}}{-2}] [x = \frac{2 \pm \sqrt{24}}{-2}] [x = \frac{2 \pm 2\sqrt{6}}{-2}]

So the x-intercepts are:

[x_1 = 1 - \sqrt{6}] [x_2 = 1 + \sqrt{6}]

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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.