Question

What is the axis of symmetry and vertex for the graph `y = -x^2 +4x + 3`?

Answer 1

We are going to use the expression to find the vertex of a parabola.

Explanation:

First of all, let us graph the curve:

graph{-x^2+4x+3 [-10, 10, -10, 10]}

This curve is a parabola, because of the form of its equation:

`y ~ x^2`

To find the vertex of a parabola, `(x_v, y_v)`, we must solve the expression:

`x_v= -b/{2a}`

where `a` and `b` are the coefficients of `x^2` and `x`, if we write parabola as it follows:

`y = ax^2+ bx + c`

So, in our case:

`x_v = - 4/{2*(-1)} = 2`

This gives us the axis of the parabola: `x=2` is the axis of symmetry.

Now, let us calculate the value of `y_v` by substituting `x_v` on parabola expression:

`y_v= - x_v^2 + 4 x_v + 3 = - 2^2+4 cdot 2 + 3 = 7`

So vertex is: `(2,7)`.