Equivalent Forms of Quadratic Equations

Question

A quadratic function that crosses the origin has a line of symmetry at x = 3, a root of 6, and a range of [-9, ∞).Write an equation in vertex form that represents the quadratic function

Write an equation in intercept form that represents the quadratic function

Write an equation in standard form that represents the quadratic function

Answer 1

y = a(x - 3)² - 9

Since 6 is a root, a(6-3)² - 9 = 0

Solve for a

Answer 2

roots of 6 and an unknown 2nd root d

means factors of x-6 and x-d

y = (x-6)(x-d)

y = x^2 - (6+d)x +6d

y' = 2x -6 -d = 0

2x = 6 + d

x = (6+d)/2 = 3

6+d = 6

d = 0

y = x^2 -6x is the quadratic with root x=6 and axis of symmetry x= 3

x = 6/2 + or - sqr(36-4(1)(0)) /2

= 3 + or - sqr36/2

= 3+ or - 3

= 0 or 6 = the two roots

but there's two quadratic equations

(x-0)(x-6) = 0 is in intercept form with intercepts 0 and 6

x^2 -6x = 0 is the quadratic in standard form for an upward opening parabola

but there's a 2nd quadratic, a downward opening parabola

-x^2 + 6x = 0 in standard form

(x-0)(-x +6) = 0 in intercept form with the same intercepts 0 and 6

both quadratics have a root of 6 and axis of symmetry of 3

range of y>-9 fits the 1st quadratic y = x^2 -6x as the vertex or minimum point is (3, -9)

but not the 2nd quadratic y=-x^2 +6x which has a range of y < 9

so the answer is y = f(x) = x^2 -6x = (x-0)(x-6)

x^2 -6x = 0 for standard form

(x-0)(x+6) = 0 or (x-0)(x - -6) = 0 for intercept form