Question

H(x)=-x^2+6x. So what is the value of h(2)

Answer 1

✩ Answer:

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✩ Step-by-step explanation:

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✺ Quadratic polynomials can be factored using the transformation
`ax^2+bx+c=a(x-x_(1))(x-x_(2) )`, where
`x_(1)` and
`x_(2)` are the solutions of the quadratic equation
`ax^2+bx+c=0`:

• 
  `-x^2+6x=0`

✺ All equations of the form
`ax^2+bx+c=0` can be solved using the quadratic formula:

• 
  `-b=\frac{+}\\√(b^2-4ac)\\~~~~~~~~~~~~~2a`

✺ The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction:

• 
  `x=\frac{\sqrt{-6\frac{+}\\√(6^2)}}{2(-1)}`

✺ Take the square root of
`6^2`:

• 
  `x=\frac{\sqrt{-6\frac{+}\\{6}}}{2(-1)}`

✺ Multiply
`2` times
`-1`:

• 
  `x=\frac{\sqrt{-6\frac{+}\\{6}}}{-2}`

✺ Now solve the equation
`x=\frac{\sqrt{-6\frac{+}\\{6}}}{-2}` when ± is plus. Add
`-6` to
`6`:

• 
  `x=(0)/(-2)`

✺ Divide
`0` by
`-2`:

• 
  `x=0`

-OR-

✺ Now solve the equation
`x=\frac{\sqrt{-6\frac{+}\\{6}}}{-2}` when ± is minus. Subtract
`6` from
`-6`:

• 
  `x=(-12)/(-2)`

✺ Divide
`-12` by
`-2`:

• 
  `x=6`

✺ Optional : Factor the original expression using
`ax^2+bx+c=a(x-x_(1))(x-x_(2) )`. Substitute
`0` for
`x_(1)` and
`6` for
`x_(2)`:

• 
  `-x^2+6x=-x(x-6)`

✩ Answer:

✺ Factored Form:
`x(x-6)`

✺ Exact Form:
`x=6`

✺ Graph Point Form:
`x=(6,0)`

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