Question

How do you find the axis of symmetry, vertex and x intercepts for `y=-3x^2+6x+1`?

Answer 1

`"vertex": (1, 4)`, `"axis of symmetry": x = 1`
`x-"intercepts": (1-(2sqrt(3))/3, 0), (1+(2sqrt(3))/3, 0)`

Explanation:

To find the vertex and axis of symmetry:
Put the equation in standard form: `y = a(x-h)^2 + k`, where axis of symmetry :`x = h` and `"vertex" = (h, k)`

First factor the `x` terms: `y = -3(x^2 -2x) + 1`

Use completing of the square:
1. Multiply the `x`-term coefficient by `1/2` to get the completed square constant: `1/2 * -2 = -1`
2. Write the completed square: `(x -1)^2 = x^2 - 2x + 1`
3. Add the `x^2`coefficient: `-3(x -1)^2 = -3x^2 +6x -3`.
4. Notice from step 3, an extra constant was added when completing the square: `-3`. This will need to be subtracted to keep the equation the same as the original equation: `y = -3(x -1)^2 + 1 -(-3)`
5. `y = -3(x -1)^2 + 4`

`"vertex": (1, 4)`, `"axis of symmetry": x = 1`

To find `x`-intercepts:
Use the quadratic equation since factoring isn't simple. Put the equation in the form: `Ax^2 + Bx +C = 0`

`x = (-B +- sqrt(B^2 - 4AC))/(2A) = (-6 +- sqrt(36-4(-3)(1)))/(2(-3))`

`x = (-6+- sqrt(48))/(-6) = 1 +- (4sqrt(3))/-6 = 1 +-(2sqrt(3))/-3`

`x-"intercepts": (1-(2sqrt(3))/3, 0), (1+(2sqrt(3))/3, 0)`

Answer 2

Please see the explanation.

Explanation:

When given a quadratic of the form: `y = ax^2+bx+c`

The equation for axis of symmetry is: `x=-b/(2a)`

The x coordinate of the vertex, h, is the same as the axis of symmetry:

`h = -b/(2a)`

The y coordinate of the vertex, k, is the function evaluated at h:

`k = y(h)`

The x intercepts can be found by factoring or by using the quadratic formula:

`x = (-b-sqrt(b^2-4(a)(c)))/(2a)` and `x = (-b+sqrt(b^2-4(a)(c)))/(2a)`

Now, to the given equation: `y = -3x^2+6x+1`

Please observe that `a = -3, b = 6, and c = 1`

The computation for the axis of symmetry is as follows:

`x = -b/(2a)`

`x = -6/(2(-3))`

`x = 1 larr` the equation for the axis of symmetry.

The x coordinate of the vertex, h, is the same as the axis of symmetry:

`h = 1`

The y coordinate of the vertex, k, is the function evaluated at h:

`k = y(1)= -3(1)^2+6(1)+1`

`k = 4`

The vertex is the point `(1,4)`

As a prelude to finding the x intercepts, compute `b^2-4(a)(c)`:

`6^2-4(-3)(1) = 36+12 = 48`

Because 48 is a positive real number, we know that there are two distinct roots and, because 48 is not a perfect square, we know that the quadratic will not factor.

Therefore, we find x intercepts using the quadratic formula:

`x = (-b-sqrt(b^2-4(a)(c)))/(2a)` and `x = (-b+sqrt(b^2-4(a)(c)))/(2a)`

`x= (-6-sqrt(48))/(2(-3))` and `x= (-6+sqrt(48))/(2(-3))`

`x= 1-sqrt(48/36)` and `x= 1+sqrt(48/36)`

`x= 1-sqrt(4/3)` and `x= 1+sqrt(4/3)larr` x intercepts