Question

k(x) = -x^2-2x+3Solve with:k(sq root 2)k(a+2)k(-x)k(x^2)The negatives are giving me a problem. Thank you

Answer 1

To answer this question, we need to evaluate the function:

$k(x)=-x^2-2x+3$

Evaluating the function for x = √2

We can proceed as follows:

$k(\sqrt[]{2})=-(\sqrt[]{2})^2-2(\sqrt[]{2})+3=-(2^{\frac{1}{2}})^2-2\sqrt[]{2}+3$$k(\sqrt[]{2})=-(2^{\frac{2}{2}})-2\sqrt[]{2}+3=-(2)-2\sqrt[]{2}+3=-2+3-2\sqrt[]{2}$$k(\sqrt[]{2})=1-2\sqrt[]{2}$

We need to remember above that:

$\sqrt[]{n}=n^{\frac{1}{2}}$

Evaluating the function for x = a + 2

We can proceed as follows:

$k(a+2)=-(a+2)^2-2(a+2)+3$$(a+2)^2=a^2+2(a)(2)+2^2=a^2+4a+4$

Then, we have:

$k(a+2)=-(a^2+4a+4)-2(a)-2(2)+3$$k(a+2)=-a^2-4a-4-2a-4+3$

Now, we need to add like terms as follows:

$k(a+2)=-a^2-4a-2a-4-4+3$$k(a+2)=-a^2-6a-8+3\Rightarrow k(a+2)=-a^2-6a-5$

Therefore

$k(a+2)=-a^2-6a-5$

Evaluating the function for x = -x

We have:

$k(-x)=-(-x)^2-2(-x)+3$$(-x)^2=(-x)(-x)=x^2$$k(-x)=-x^2+2x+3$

Evaluating the function for x = x²

We can proceed similarly:

$k(x^2)=-(x^2)^2-2(x^2)+3=-(x^4)-2x^2+3$

Therefore, we have:

$k(x^2)=-x^4-2x^2+3$

In summary, we have that:

$k(\sqrt[]{2})=1-2\sqrt[]{2}\approx-1.82842712475$$k(a+2)=-a^2-6a-5$$k(-x)=-x^2+2x+3$$k(x^2)=-x^4-2x^2+3$