Expression 5 has the term \(\displaystyle -3\sqrt{x}\), which violates the definition of a polynomial.  The exponent must be a whole number.

When finding the x-intercepts for a function, this is where the function crosses the x-axis, which means that \(\displaystyle y\) or \(\displaystyle \ f(x)\) must equal zero. 

So we first set \(\displaystyle f(x)=0\) which gives us: 

\(\displaystyle 0=x^{3}+3x^{2}-4x-12\)

Now in order to solve this equation, we must break down the polynomial using the "Factor by Grouping" Method. 

To "Factor by Grouping" you must put the polynomial in standard form and then group into to pairs of binomials. 

\(\displaystyle (x^{3}+3x^{2})(-4x-12)\) 

After doing this, one can see that there is a common factor in each group. 

\(\displaystyle x^{2}(x+3) -4(x+3)\)

When an \(\displaystyle x^{2}\) is taken out of the first pair we are left with \(\displaystyle (x+3)\) and, 

when a \(\displaystyle -4\) is taken out of the second pair we are left with \(\displaystyle (x+3)\) again. 

The goal is to make each the same and we now have two \(\displaystyle (x+3)'s\). 

This is now a common factor on this side of the equation, so we can take out the common factor \(\displaystyle (x+3)\) and we get ths result. 

\(\displaystyle (x+3)(x^{^{2}}-4)\)

We can now find the x-intercepts by remembering that we origianilly set this all equal to 0. 

\(\displaystyle (x+3)(x^{^{2}}-4)=0\)

In order for this product to equal zero, either the first or second parentheses needs to equal zero, so we set each equal to zero and solve. 

\(\displaystyle x+3=0\and\ x^{2}-4=0\)

\(\displaystyle x=-3\) and \(\displaystyle x^{2}=4\).

After taking the square root of both sides for \(\displaystyle x^{2}=4\) you get \(\displaystyle x=\pm2\).

Substitute 5y in for every x:

\(\displaystyle -2(5y)^2+3(5y)-12\).

Simplify:

Square the first term: \(\displaystyle -2(25y^2)+3(5y)-12\)

Distribute the coefficients: \(\displaystyle -50y^2+15y-12\)

Substitute \(\displaystyle (2a-1)\) for \(\displaystyle x\) in the original equation:

\(\displaystyle -2(2a-1)^2+3(2a-1)-12\).

Use FOIL or the Square of a Binomial Rule to find \(\displaystyle (2a-1)^2=(4a^2-4a-1)\).

Recall that FOIL stands for the multiplication between the First components in both binomials followed by the Outer components, then the Inner components, and lastly the Last components.

Then, Distribute: \(\displaystyle (-8a^2+8a-2)+(6a-3)-12\).

Combine like terms to simplify:

\(\displaystyle -8a^2+14a-5-12\)

\(\displaystyle -8a^2+14a-17\)

To find \(\displaystyle b(3x-1)\), substitute \(\displaystyle (3x-1)\) for \(\displaystyle x\) in the original equation:

\(\displaystyle 2(3x-1)^2+(3x-1)-1\).

Use FOIL or the Square of Binomial Rule to find \(\displaystyle (3x-1)^2=(9x^2-6x+1)\).

Recall that FOIL stands for the multiplication between the First components in both binomials followed by the Outer components, then the Inner components, and lastly the Last components.

Therefore, \(\displaystyle b(3x-1)=2(9x^2-6x+1)+(3x-1)-1\). 

You can then simplify the equation.

Distribute the multiplier: \(\displaystyle 18x^2-12x+2+3x-1-1\)

Combine like terms: \(\displaystyle b(3x-1)=18x^2-9x\).

To find \(\displaystyle 3b(x)\), distribute 3 throughout the equation to get:

\(\displaystyle 3b(x)=3(2x^2+x-1)=\)\(\displaystyle 6x^2+3x-3\). 

Subtract the two expressions:

\(\displaystyle (18x^2-9x)-(6x^2+3x-3)\)

\(\displaystyle =12x^2-12x+3.\)

For \(\displaystyle f(5a-2)\), substitue\(\displaystyle (5a-2)\) for \(\displaystyle x\):

\(\displaystyle (5a-2)^2-5(5a-2)+8\).

Use FOIL or square of a binomial to find \(\displaystyle (5a-2)^2=25a^2-20a+4\).

Recall that FOIL stands for the multiplication between the First components in both binomials followed by the Outer components, then the Inner components, and lastly the Last components.

Therefore, \(\displaystyle f(5a-2)=(25a^2-20a+4)-5(5a-2)+8\)

Distribute and combine like terms to simplify:

\(\displaystyle (25a^2-20a+4)-(25a-10)+8\)

\(\displaystyle f(5a-2)=25a^2-45a+22\).

For \(\displaystyle 3f(2a)\), first substitute \(\displaystyle (2a)\) for \(\displaystyle x\):

\(\displaystyle 2a^2-5(2a)+8\).

Multiply the entire expression by 3:

\(\displaystyle 3f(2a)=12a^2-30a+24\).

Add both expressions:

\(\displaystyle (25a^2-45a+22)+(12a^2-30a+24)\)

\(\displaystyle =37a^2-75a+46\)

To solve this problem, plug in 2p for x in the function: \(\displaystyle 2(2p)^3\). Then, simplify: \(\displaystyle 2(8p^3)=16p^3\).

The question asks us to put the expression of \(\displaystyle g(x)\) into the expression for \(\displaystyle f(x)\) anyplace there is an \(\displaystyle x\):

\(\displaystyle f(g(x))=(-3x^{-\frac{1}{2}})^2+1\)

The 2nd power needs to be distributed to both the \(\displaystyle -3\) and \(\displaystyle x^{-\frac{1}{2}}\). The first term then becomes:

\(\displaystyle 9x^{-1} = \frac{9}{x}\)

The final answer is then \(\displaystyle \frac{9}{x}+1\)

To solve this problem, simply plug in 1 wherever you see x.

\(\displaystyle -2(1^3)+1-4=-2+1-4=-5\).

Therefore,

\(\displaystyle f(1)=-5\).

To determine the end behavior for a function, we must look at the degree and the sign associated with the function. For this function: 

\(\displaystyle f(x)=x^6-7x^3+2x^2-5\)

The degree is 6 (the highest power) which is an even number, and the sign is positive (the sign associated with the leading coefficient, which in this case is positive 1). 

Taken together, we can see that we have an "even positive" function. Even functions always go the same direction at both extreme ends. Think of a parabola, both sides either both go up or both go down - this is the same for all even functions. 

Since this is an even positive; both sides will approach positive infinity. 

We express this mathematically when we say that as x approaches negative infinity (left side) the function will approach positive infinity:

\(\displaystyle As\ x\rightarrow\ -\infty; f(x)\rightarrow\infty;\)

....and as x approaches positive infinity (right side) the function again approaches positive infinity. 

\(\displaystyle As\ x\rightarrow\infty; f(x)\rightarrow\infty;\)