How do you find the important points to graph f(x)=x^2+2x?

1 Answer

Find the Vertex, X-intercepts, y-intercept

Explanation:

From the given y=x^2+2x

the format of f(x)=y=ax^2+bx+c

The Vertex (h, k)

with a=1and b=2 and c=0

h=-b/(2a)=-2/(2*1)=-1

k=c-b^2/(4a)=0-2^2/(4*1)=-4/4=-1

Vertex (h, k)=(-1, -1)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The x-intercepts (x_1, 0) and (x_2, 0)

From the given equation y=x^2+2x, set y=0,then solve for x values

0=x^2+2x
0=x(x+2)

then x_1=0 and x_2=-2
The x-intercepts (x_1, 0) and (x_2, 0) are (0, 0),and (-2, 0)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The y-intercept (0, y_1)

From the given equation y=x^2+2x, set x=0 then solve for the y value

y=x^2+2x

y=0^2+2*(0)

y=0 when x=0

The y-intercept (0, y_1)=(0, 0)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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I hope the explanation is useful....God bless