We did this qs in class so we change the equation into the vertex form
Here is the solution.
Y=3x^2+30x+41
y=3(x^2+10x)+41
Y=(x^2+10x+25-25)+41
Y=3(x+5)^2-34
I understand everything expected I don't know how we get -34, can you explain please?
On the third line you have the +25 and -25
they got rid of the -25
To do this they removed it from the bracket to the outside of the function
But since it was in the bracket you cannot forget to multiply by the 3 that was on the outside
This made the -25 a -75....and when you add the -75 and 41 you get -34
Certainly! To understand how we obtain the constant term of -34 in the vertex form of the equation, let's break down the steps:
1. Start with the given quadratic equation: Y = 3x^2 + 30x + 41.
2. Group the first two terms, 3x^2 and 30x: y = 3(x^2 + 10x) + 41.
3. Notice that the expression inside the parentheses, x^2 + 10x, can be rewritten as a perfect square by adding (10/2)^2 = 25 to complete the square. This maintains the equivalence of the equation.
4. Rewrite the equation with the completed square: y = 3(x^2 + 10x + 25 - 25) + 41.
5. Factor the perfect square trinomial (x^2 + 10x + 25) as (x + 5)^2: y = 3((x + 5)^2 - 25) + 41.
6. Distribute the 3 into the parentheses: y = 3(x + 5)^2 - 75 + 41.
7. Simplify the constants: y = 3(x + 5)^2 - 34.
Hence, the constant term of -34 in the vertex form of the equation is obtained by combining the constants -75 and 41.