The tutor continues about converting from standard to vertex form.

Yesterday’s post opened the discussion about changing a quadratic function from standard form

y=ax² + bx + c

to vertex form

y=a(x-p)² + q

That post focused on the case where a=1:

y=x² +bx +c

Today, the case where a≠1:

Example:   Convert y=3x² + 24x – 5 to vertex form.

Step 1: Like before, rewrite the equation with a space before the constant term:

y=3x²+ 24x      -5

Step 2: Factor the coefficient of x² from both variable terms:

y=3(x² +8x    )      -5

Step 3: Inside the brackets, complete the square (see my post here for an introduction): take half the coefficient of the x term, square it, then add it inside the brackets.

In this case, the coefficient of x is 8. We take 4, square it to get 16, and then write that in the brackets.

y=3(x² + 8x + 16) -5

Step 4: Very important: Realize that when you added the number in the brackets, you really added that number times the number in front. You need to subtract that product on the outside, to equalize.

In this case, we added 16 in the brackets; therefore, we really added 3(16)=48. To equalize, we must subtract 48 from the outside:

y=3(x² +8x +16) -5 -48

Step 5: Now realize that the trinomial in the brackets is a perfect square. (Once again, see yesterday’s post.)

y=3(x² +8x +16) – 53 becomes y=3(x+4)² -53

Apparently, y=3x² +24x -5, in vertex form, is y=3(x+4)² -53. The vertex is (-4,-53). (See my post here about identifying the vertex.)

While this shows the general case of converting from standard form to vertex form, it’s an easy example. I’ll cover some more difficult ones in future posts.

HTH:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.