Question

how do I factorise y= -2(x-5)^2 +9 so I get x intercepts

Answer 1

Answer:

Recall that:

$a^2-b^2=(a+b)(a-b).$

Now, notice that:

$\begin{gathered} -2(x-5)^2+9=-2((x-5)^2-\frac{9}{2})=-2((x-5)^2-\frac{3^2}{\sqrt{2}^2}) \\ =-2((x-5)^2-(\frac{3}{\sqrt2})^2)=-2((x-5)^2-(\frac{3\sqrt{2}}{2})^2). \end{gathered}$

Using the first property in the answering tab we get:

$(x-5)^2-(\frac{3\sqrt{2}}{2})^2=(x-5-\frac{3\sqrt{2}}{2})(x-5+\frac{3\sqrt{2}}{2})$

Therefore we can rewrite the given equation as follows:

$y=-2(x-5-\frac{3\sqrt{2}}{2})(x-5+\frac{3\sqrt{2}}{2}).$

Therefore the x-intercepts of the given equation are:

$\begin{gathered} x=-(-5-\frac{3\sqrt{2}}{2})=5+\frac{3\sqrt{2}}{2} \\ and \\ x=-(-5+\frac{3\sqrt{2}}{2})=5-\frac{3\sqrt{2}}{2}. \end{gathered}$