writ the equation of the axis of symmetry and the coordinates of the vertex of each quadratic function?

Question

please can you explain me how to get the answer.

y=-x+4x+5

sorry, i made a mistake, it was y=-x^2+4x+5

Answer 1

y = -x^2 + 4x + 5

x = (-b/2a)
x = (-4)/(2(-1))
x = (-4)/(-2)
x = 2

y = -(2)^2 + 4(2) + 5
y = -4 + 8 + 5
y = 4 + 5
y = 9

Vertex : (2,9)
Axis of symmetry is x = 2

Answer 2

The problem should be phrased as

y=-x^2+4x+5

Your problem is simply a graphical application of the "complete the square" exercise. Let's complete the square here:

-x^2+4x+5=
-(x^2-4x-5)=
-(x^2-4x-4+4-5)=
-((x^2-4x+4)-1)=
-(x-2)^2+1=y

From here, you first see that x=2 is the axis of symmetry of this graph, which as you know is a parabola.

Vertex is the coordinates of the intersection of the graph with its axis of symmetry. That means that x=2 from above. Plugging in into y gives y=0+1=1.

The coordinates are thus V(2;1)

Answer 3

To find the axis of symmetry, use the formula x = -b/2a.
In this case, that would be x = -4/2(-1), simplified to x = 2.
That is the equation for the axis of symmetry, x = 2.
Then, by substitution, y = -(2)^2 + 4(2) + 5
y = -4 + 8 + 5
y = 9

The vertex is on (2, 9).