does anyone know how to find axis of symmetry and vertex?

Question

5x^2 + 8x + 13

Answer 1

Axis of symmetry for a parabola is the line passing through the focus and vertex of a parabola,
So, we first find the vertex

Vertex of a Parabola is the point at which a parabola makes its sharpest turn
For a parabola y = a*x^2 + bx + c

If you express your parabola equation y = a*x^2 + bx + c in form
(x-h)^2 = 4p*(y-k) Then, the vertex is (h,k) and focus is (h, k + p).

For your parabola y = 5x^2 + 8x + 13
=> y - 13 = 5x^2 + 8x
=> (y - 13)/5 = x^2 + (8/5)x + 16/25 - 16/25
=> (y - 13)/5 = (x^2 + (8/5)x + 16/25) - 16/25
=> (y - 13)/5 = (x^2 + (8/5)x + 16/25) - 16/25
=> (y - 13)/5 = (x^2 + (4/5)) ^2 - 16/25
=> (y - 13)/5 + (16/25)= (x^2 + (4/5)) ^2
=> (y - (49/5))/5 = (x^2 + (4/5)) ^2
=> (x^2 + (4/5)) ^2 = (y - (49/5))/5
Comparing with (x-h)^2 = 4p*(y-k), we get
h = (-4/5), k = 49/5, 4p = 1/5 and p = 1/20

Thus, focus is (h, k + p) = (-4/5 , 49/5 +1/20)
= (-4/5 , 49/5 +1/20)
= (-4/5 , 54/5)

And vertex is (h,k) = (-4/5 , 1/20)

So, the axis of symmetry is line passing through (-4/5 , 49/5) and (-4/5 , 54/5).

Its clear that the only line passing through these points is x = -4/5.

Answer 2

The line of symmetry will always be at x=-b/2a
This gives x - 8/(2*5) = -4/5.

Since a = 5 is positive this represents a minimum and the parabola is concave down (like a bowl).

Put x in the original equation and get y= 5(-4/5)^2 +8(-4/5) + 13
= 16/5- 32/5 +13= 13-16/5 = 65-16/5= 49/5

Thus the vertex is at (-4/5, 49/5)

Answer 3

Y' = 10x +8 (Slope)
10X + 8 = 0
X = -4/5
symmetric about vertical line at X = -4/5

5(-4/5)^2 + 8(-4/5) + 13 = -9.8
Vertex: (-4/5, -9.8)