Question

What is the axis of symmetry and vertex for the graph `y=3x^2+12x-2`?

Answer 1

Axis of symmetry: `x = -2`
Vertex: `(-2, -14)`

Explanation:

This equation `y = 3x^2 + 12x - 2` is in standard form, or `ax^2 + bx + c`.

To find the axis of symmetry, we do `x = -b/(2a)`.

We know that `a = 3` and `b = 12`, so we plug them into the equation.
`x = -12/(2(3))`

`x = -12/6`

`x = -2`

So the axis of symmetry is `x = -2`.

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Now we want to find the vertex. The `x`-coordinate of the vertex is the same as the axis of symmetry. So the `x`-coordinate of the vertex is `-2`.
To find the `y`-coordinate of the vertex, we just plug in the `x` value into the original equation:
`y = 3(-2)^2 + 12(-2) - 2`

`y = 3(4) - 24 - 2`

`y = 12 - 26`

`y = -14`

So the vertex is `(-2, -14)`.

To visualize this, here is a graph of this equation:

Hope this helps!

Answer 2

Axis of Symmetry is the line `color(blue)(x=-2`

Vertex is at: `color(blue)((-2, -14).`It is a minimum.

Explanation:

Given:

`color(red)(y=f(x) = 3x^2+12x-2`

We use the Quadratic formula to find the Solutions:

`color(blue)(x_1, x_2=(-b+-sqrt(b^2-4ac))/(2a)`

Let us look at `color(red)(f(x)`

We observe that `color(blue)(a=3; b=12; and c=(-2)`

Substitute these values in our Quadratic formula:

We know that our discriminant `b^2-4ac` is greater than zero.

`color(blue)(x_1, x_2=[-12+-sqrt[12^2-4(3)(-2)])/(2(3))`

Hence, we have two real roots.

`x_1, x_2=[-12+-sqrt([144+24]])/(6)`

`x_1, x_2=[-12+-sqrt(168])/(6)`

`x_1, x_2=[-12+-sqrt(4*42])/(6)`

`x_1, x_2=[-12+-sqrt(4)*sqrt(42])/(6)`

`x_1, x_2=[-12+-2*sqrt(42])/(6)`

`x_1, x_2=[-12]/6+-(2*sqrt(42])/(6)`

`x_1, x_2=-2+-(cancel 2*sqrt(42])/(cancel 6 color(red)3)`

`x_1, x_2=-2+sqrt(42)/3, -2-sqrt(42)/3`

Using a calculator, we can simplify and get the values:

`color(blue)(x_1=0.160247, x_2=-4.16025`

Hence, our x-intercepts are: `color(green)((0.16,0),(-4.16,0)`

To find the Vertex,

we can use the formula: `color(blue)((-b))/color(blue)((2a)`

Vertex: `-12/(2(3)`

`rArr -12/6=-2`

This is our x-coordinate value of our Vertex.

To find the y-coordinate value of our Vertex:

Substitute the value of `color(blue)(x=-2` in

`color(red)(y = 3x^2+12x-2`

`y = 3(-2)^2+12(-2)-2`

`y = 3(4)-24-2`

`y = 12-24-2 = 14`

Vertex is at: `color(blue)((-2, -14)`

The coefficient of the `color(green)(x^2` term is Positive and hence, our Parabola Opens Upward, and it has a minimum. Please refer to the image of the graph below to verify our solutions:

The Axis of symmetry of a parabola is a vertical line that divides the parabola into two congruent halves.

The Axis of Symmetry always passes through the Vertex of the Parabola. The `x` coordinate of the vertex is the equation of the Axis of Symmetry of the Parabola.

Axis of Symmetry is the line `color(blue)(x=-2`