$f(x)=x^3+x^2-6x\\\\text{The zeros:}\\\\f(x)=0\to x^3+x^2-6x=0\\\\x(x^2+x-6)=0\\\\x(x^2+3x-2x-6)=0\\\\x[x(x+3)-2(x+3)]=0\\\\x(x+3)(x-2)=0\iff x=0\ \vee\ x+3=0\ \vee\ x-2=0\\\\x=0\ \vee\ x=-3\ \vee\ x=2$

Answer: -3; 0 and 2

6 0

3 years ago

How is the graph of y = (x + 1)^2- 5 transformed from the graph of y=x^2?

Ainat [17]

$\text{Hi there! :D}$

$\boxed{\text{There is a horizontal shift of 1 unit left and 5 units down.}}$

$\text{Recall that the transformations form of a quadratic function is:}\\\\y = \pm a(b(x-h))+k \text{ where:}\\\\a = \text{ vertical stretch/compression}\\\\b = \text{ horizontal stretch/compression}\\\\h = \text{ shift left/right}\\\\k = \text{ shift up/down}\\\\\text{Therefore, in this instance:}\\\\h = -1\\\\k = -5, \text{ so:}\\\\\text{There is a horizontal shift of 1 unit left and 5 units down.}$