I have got help from Damon on this question but all I need is help on integrating the answer he gave me into my original equation:

y=-15/20.25(x+1.5)^2+0

Here's Damon's help: (x+1.5)(x+1.5) = x^2 + 1.5 x + 1.5 x + (1.5*1.5)

= x^2 + 3 x + 2.25

So would I put this into my original quadratic equation?

Please and thank you as the help would be much appreciated!

Yes, you can integrate the expression Damon provided into your original equation. Let's break it down step by step:

Original equation: y = -15/20.25(x+1.5)^2 + 0

Step 1: Apply the substitution into your original equation. Replace (x+1.5)^2 with the expanded expression provided by Damon:

y = -15/20.25 * (x^2 + 3x + 2.25) + 0

Step 2: Simplify the expression by multiplying the coefficient (-15/20.25) with each term inside the parentheses:

y = (-15/20.25) * x^2 + (-15/20.25) * 3x + (-15/20.25) * 2.25

Simplifying further gives:

y = (-15/20.25) * x^2 - (45/20.25) * x - (33.75/20.25)

You can simplify the coefficients if desired, but this is the final form of the equation obtained by integrating Damon's help into your original quadratic equation.

Hope this helps! Let me know if you have any further questions.