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:

• $-b=\frac{+}\\\sqrt{b^2-4ac}\\~~~~~~~~~~~~~2a$

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

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

✺ 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 $6$:

• $x=\frac{0}{-2}$

✺ Divide $0$ by $-2$:

• $x=0$

       -OR-

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

• $x=\frac{-12}{-2}$

✺ 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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