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

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Answer 1 · 2016-03-25T03:04:26

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

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)#.