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

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]}

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)#.