What is the the vertex of #y=(x -1)^2 + 2x +16#?

Expand the brackets first

To find the vertex of the quadratic function ( y = (x - 1)^2 + 2x + 16 ), first, we rewrite the equation in the standard form ( y = ax^2 + bx + c ). Then, we use the formula ( x = -\frac{b}{2a} ) to find the x-coordinate of the vertex. Finally, we substitute this x-coordinate into the original equation to find the y-coordinate of the vertex.

First, let's rewrite the equation in standard form: [ y = (x - 1)^2 + 2x + 16 ] [ y = (x^2 - 2x + 1) + 2x + 16 ] [ y = x^2 - 2x + 1 + 2x + 16 ] [ y = x^2 + 16 ]

Comparing this equation with ( y = ax^2 + bx + c ), we see that ( a = 1 ), ( b = 0 ), and ( c = 16 ).

Now, using the formula ( x = -\frac{b}{2a} ), we find: [ x = -\frac{0}{2(1)} ] [ x = 0 ]

Now that we have the x-coordinate of the vertex, let's find the y-coordinate by substituting ( x = 0 ) into the original equation: [ y = (0 - 1)^2 + 2(0) + 16 ] [ y = 1 + 0 + 16 ] [ y = 17 ]

So, the vertex of the function is at ( (0, 17) ).