Question

How do you find the vertex of `y=-x^2+4x+12`?

Answer 1

Let's look at the quadratic formula:

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

Knowing parabolas are symmetric around the vertex, we can infer that the axis lies on the vertical line defined by the non-variant part of the quadratic formula (a.k.a the one not affected by the `+-`)

`"Non variant part: " (-b)/(2a)`

We know the result of that (after plugging `a = -1, b = 4`) will be the x-value of the vertex. To find the y-value, we simply plug in the x-value into the function to see its value.

So, generalising, the vertex of a quadratic function `f(x)` is always:

`V((-b)/(2a), f((-b)/(2a)))`

In this case it's `V(2,16)`

graph{-x^2 + 4x +12 [-16.81, 19.23, -0.73, 17.29]}

Answer 2

`(2,16)`

Explanation:

Well, firstly find the axis of symmetry for the parabola. The formula for this axis of symmetry is `-b/(2a)`.

Now you may not know what `a` and `b` stand for, but this `ax^2 + bx +c` formula that you have seen this is the general equation of a parabola.

Notice the `a` in front of the `x^2` that is a constant that is your `a` in the axis of symmetry equation `-b/(2a)`. Similarly, the `b` in front of the `x` in your general parabola equation is the `b` in your axis of symmetry equation `-b/(2a)`.

So sub in your values for `a` and `b` and you will get

`(-4)/-2 = 2`

Thus your axis of symmetry is `x = 2`.

To find your vertex, sub this `2` into your main equation which was

`y = -x^2 + 4x + 12`

So

`-(2)^2 + 4 * 2 + 12 = 16`

Finally, your vertex is at `(2,16)`.