For the function y = x^2 - 2x - 5, determine
a) Its Vertex
b) The x-intercepts in simplified form
2 Answers | Add Yours
First, we recognize this is a quadratic with a U-shaped graph. There is a formula that exists for the x-coordinate for the vertex:
x = -b/(2a)
Where a and b come from the equation y = ax^2 + bx + c. Here, a = 1, b = -2. So, we have:
x = -(-2)/(2*1) = -1
Again, that is the x coordinate of the vertex. Whenever we have any x, to find the y, we can plug it into the original equation. So:
y = (-1)^2 - 2(-1) - 5 = 1 + 2 - 5 = -2
So, the vertex is (-1,-2)
For all x-intercepts, y = 0. So, for this, it would be exactly like solving the equation:
0 = x^2 - 2x - 5
This doesn't factor. So, we have to use the quadratic formula or complete the square. Doing complete the square, first, add 5 to each side. So:
5 = x^2 - 2x
Then, take half of b, -2/2 = -1. Then, square that result, (-1)^2 = 1. Add that result to both sides.
5+1 = x^2 - 2x + 1
The right side factors now:
6 = (x-1)^2
Square root each side, then add 1 to each side, we have:
x = 1 +- sqrt(6)
So, the x intercepts are (1 + sqrt 6, 0) and (1-sqrt 6, 0).
Good luck, Kristen. I hope this helps.
If the equation of a parabola y = ax^2 + bx + c is written in the form y = a(x - h)^2 + k, the vertex of the parabola is the point (h, k).
For the parabola y = x^2 - 2x - 5, to determine the vertex write the equation in the form above:
y = x^2 - 2x - 5
y = x^2 - 2x + 1 - 6
y = (x - 1)^2 - 6
This gives the vertex as (1, -6)
At the points where the parabola intersects the x-axis the value of y is 0.
Solve the equation x^2 - 2x - 5 = 0
x^2 - 2x + 1 - 6 = 0
(x - 1)^2 = 6
`x - 1 = +- sqrt6`
`x = 1 +- sqrt6`
The x-intercepts of the parabola are `x +- sqrt6`
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes