give an example of a quadratic function with real zero x=-3 that is increasing on (-5,-3)?

Let f(x)=ax^2+bx+c, then
f '(x)=2ax+b
=>
9a-3b+c=0--------(1)
25a-5b+c=-3-----(2)
2a(-5)+b>0
=>
b>10a-------------(3)
(2)-(1)=>
16a-2b=-3
=>
8a-b=-1.5
taking b=11a, then
8a-11a=-1.5
=>
a=0.5
=>
b=5.5
From (1), get
c=3(5.5)-9(0.5)=12
Thus,
f(x)=0.5x^2+5.5x+12
is one of the examples.

Check:
f(x)=0.5x^2+5.5x+12 is a
quadratic function.
f(-3)=0.5(-3)^2+5.5(-3)+12=0
=>
x=-3 is a root.
f '(-5)=2(0.5)(-5)+5.5=0.5>0
=>
f(x) is increasing at (-5, -3).

f(x) = (x-r)(x+3) = x² + (3-r)x - 3r
f'(x) = 2x + 3 - r > 0 on (-5, -3)
r < 2x+3
r < -10+3 = -7
r < -6+3 = -3

The function will be increasing on (-5,-3) if the roots are r₁ = -3, and r₂ < -7.
Choose roots:
r₁ = -3
r₂ = -8

Ans:
An example of a quadratic function will real zero x = -3 that is increasing on (-5,-3) is:
f(x) = (x+8)(x+3) = x² + 11x + 24

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A second approach that doesn't require derivative calculus would be to recognize that any concave upward parabola with a vertical axis x < -5 and a zero at x = -3 would satisfy the conditions. Since the vertex is midway between the zeros, select one zero at x = -3 and any other zero at x < -7