Please help me with this thanks and plz explainplz help

Accepted Solution

You're looking to put this into vertex form, which is:

[tex]\sf y=a(x-h)^2+k[/tex]

Where [tex]\sf (h,k)[/tex] is the vertex of the parabola(minimum).

You currently have it in standard form, which is:

[tex]\sf y=ax^2+bx+c[/tex]

To go from standard form to vertex form, we need to complete the square. But first, let's factor out the coefficient of the first term out of the first two terms:

[tex]\sf p^2-28p+250[/tex]

[tex]\sf 1(p^2-28p)+250[/tex]

Now let's complete the square inside of the parenthesis. We take half of the second term(-28p), square it, and then add it and subtract it into the parenthesis(because we have to keep the function the same). Half of -28 is -14, -14 squared is 196:

[tex]\sf 1(p^2-28p+196-196)+250[/tex]

Now the first three terms in the parenthesis are a perfect square, we can factor them:

[tex]\sf 1((p-14)^2-196)+250[/tex]

Distribute 1 into the parenthesis(anything times 1 is itself):

[tex]\sf (p-14)^2-196+250[/tex]

Combine like terms:

[tex]\boxed{\sf (p-14)^2+54}[/tex]