Converting Quadratics: Standard Form and Vertex Form

To convert from vertex form $y=3(x+2{)}^2-4$ to general form, start by expanding $(x+2{)}^2$ to get $x^2+4x+4$. Then, distribute the 3 to each term: $3(x^2+4x+4)=3x^2+12x+12$. Finally, subtract the 4 outside the brackets: $3x^2+12x+12-4$. Simplifying, we get $y=3x^2+12x+8$.

To convert from vertex form $y=4(x-1{)}^2+2$ to general form, start by expanding $(x-1{)}^2$ to get $x^2-2x+1$. Then, distribute the 4 to each term: $4(x^2-2x+1)=4x^2-8x+4$. Finally, add the 2 outside the brackets: $4x^2-8x+4+2$. Simplifying, we get $y=4x^2-8x+6$.

To convert to vertex form, we complete the square. We want to turn $x^2+4x$ into a perfect square. Take half of 4, which is 2, then square it: $2^2=4$. Now we have $y=x^2+4x+4=(x+2{)}^2$. But since we added 4, we need to subtract it to keep the expression balanced: $y=(x+2{)}^2+5-4$. So the equation becomes $y=(x+2{)}^2+1$.

First, complete the square: take half of −6 (which is −3), then square −3 to get 9. Now, rewrite the equation as $y=(x^2-6x+9)+8-9$. We added 9 inside the brackets to complete the square, so we must subtract 9 to keep the equation balanced. After simplifying, we get $y=(x-3{)}^2-1$.

First complete the square: take half of −8 (which is −4), then square −4 to get 16. Now, rewrite the equation as $y=(x^2-8x+16)+15-16$. We added 16 inside the brackets to complete the square, so we must subtract 16 to keep the equation balanced. After simplifying, we get $y=(x-4{)}^2-1$.

First, factor out the 3 from the $x^2$ and $x$ terms: $y=3(x^2-4x)+5$. Next, complete the square inside the brackets. Half of −4 is −2, and squaring −2 gives 4. Now, rewrite the equation as $y=3(x^2-4x+4)+5-3(4)$. We added 4 inside the brackets, so we subtract $3\times 4=12$ outside to keep the equation balanced. Simplifying, we get $y=3(x-2{)}^2-7$.