158k views
0 votes
a parabola with a vertex (1,5) and y-intercept (0,2) crosses the x-axis in two places. one x-intercept is at (-0.29,0) what is the other x-intercept?

2 Answers

4 votes

Answer:

(2.29,0)

Explanation:

Edmentum

User Aghilas Yakoub
by
5.2k points
2 votes

The other x intercept would be 2.29.


You can find this by first modeling the equation using vertex form: y = a(x - h)^2 + k, in which the variables stand for the following:


a = constant

h = x value of vertex

k = y value of vertex.


Since we know enough to plug in and solve for a, we can get the equation we are using.


y = a(x - h)^2 + k


First put in the vertex points into (h, k)


y = a(x - 1)^2 + 5


Now add the points we know for x and y from the y-intercept


2 = a(0 - 1)^2 + 5


And solve that for a


2 = a(0 - 1)^2 + 5

2 = a (-1)^2 + 5

2 = a + 5

-3 = a


Now we can model the equation as y = -3(x - 1)^2 + 5. Using that, we can distribute to put in standard form.


y = -3(x - 1)^2 + 5

y = -3(x^2 - 2x + 1) + 5

y = -3x^2 + 6x - 3 + 5

y = -3x^2 + 6x + 2


Now we can use the quadratic equation to solve for the remaining zero. Using the following variables:


a = -3 (attached to x^2)

b = 6 (attached to x)

c = 2 (constant)



\frac{-b +/- \sqrt{b^(2) - 4ac}}{2a}



\frac{-6 +/- \sqrt{6^(2) - 4(-3)(2)}}{2(-3)}



(-6 +/- √(36 + 24))/(-6)



(-6 +/- √(60))/(-6)



(-6 +/- 7.74)/(-6)


x = 0.29 or x = -2.29

User Dequin
by
5.8k points