Please help! Struggling with these Algebra questions!!

The width of a rectangle is 1ft less than the length. The area is 2ft^2 Find the length and the width.

Find the x and y intercepts f(x)= -x^2+2x+63

Find the x and y coordinate of the vertex, line of symmetry, the maximum or minimum value of the quadratic function and graph the function. f(x)= -2x^2+2x+5

and the last question is the same as the one above this but it's f(x)= 1 - x^2

Hello

w = l - 1
Area = 2 = l*w = l* (l-1) = l^2 - l
l^2 - l - 2 = 0
l = 0.5 +- √(0.5^2 + 2)
l = 0.5 +- 1.5
l1 = - 1 --> doen't make sense
l2 = 2 -- solution for l
w = 2/l = 1

length = 2, width = 1
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y = -x^2 + 2x + 63

x intercepts: set y = 0
-x^2+2x+63 = 0
x^2 - 2x - 63 = 0
x = 1 +- √(1^2 +63)
x = 1 +- 8
x1 = 9
x2 = -7 = x intercepts

y intercept:
set x = 0
y = 63 = y intercept
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y = -2x^2 + 2x + 5
y' = -4x + 2 = 0
x = 0.5
y = -2*0.5^2 + 2*0.5 + 5
y = 5.5

the vertex is at (0.5, 5.5)
the line of symmetry is x = 0.5
the maximum value is y of the vertex (= 5.5)
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y = 1-x^2
y' = - 2x = 0
x = 0

y = 1 - 0^2 = 1

the vertex = (0, 1)
line of symmetry = x = 0 (= y axis)
maximum = vertex = y = 1 at x = 0

Regards