Acc. Math I: Converting Between Standard Form, Vertex Form, and Intercept Form

Acc. Math I: Converting between standard form, vertex form, and intercept form

Standard form of a quadratic: y = ax2 + bx + c

where the vetex is

For example: y = 2x2 – 8x + 1 a = 2 b = -8 c = 1

Now substitute 2 in for x and solve for y: 2(2)2 – 8(2) + 1 = -7

So the vertex is (2, -7)

Vertex form of a quadratic: y = a(x – h)2 + k where the vertex is (h, k)

For example: y = -3(x – 1)2 + 4 the vertex is (1, 4)

y = 2(x + 6)2 the vertex is (-6, 0)

y = 5x2 – 1 the vertex is (0, -1)

Intercept form of a quadratic: y = a(x – p)(x – q) where (p, 0) and (q, 0) are the x-intercepts, or p and q are the roots. Also, the x-coordinate of the vertex will be half way between p and q. Plug the x-coordinate in to find the y-coordinate.

For example: y = -2(x – 4)(x + 6) the x-intercepts are (4, 0) and (-6, 0)

and the roots are 4 and -6. To find the vertex, find the half-way point between 4 and -6: (4 + -6)/2 = -1 and plug -1 in for x to find y: (-2)(-1 - 4)(-1 + 6) = 50 So the vertex is (-1, 50).

To convert from standard form to vertex for, you can either use completing the square or just find the vertex and substitute in for a, h, and k in vertex form.

Let’s practice: Convert 2x2 + 4x – 5 into vertex form using completing the square:

Using finding the vertex and substituting:

To convert from vertex form into standard form, follow order of operations. Work the exponent first followed by distribution, then add or subtract.

Let’s practice: Convert y = -2(x – 3)2 + 6 into standard form.

To convert from intercept form to standard form, multiply the binomials then distribute.

Let’s practice: Convert y = 3(x – 1)(x + 2) into standard form, then vertex form.