Question

Y f(x)=ax^2+bx+c Graph: f(x)=x^2+4x+4 I a= h=-b/2a k= Vertex= y-int= Page 8 I 8 + Search o

Answer 1

Given the parabola:

`y=x^2+4x+4`

a = 1

b = 4

c = 4

The x-coordinate of the vertex is found as follows:

`\begin{gathered} h=(-b)/(2a) \\ h=(-4)/(2\cdot1) \\ h=-2 \end{gathered}`

The x-coordinate of the vertex is found as follows:

`\begin{gathered} k=y(h)=h^2+4h+4 \\ k=(-2)^2+4(-2)+4 \\ k=4-8+4 \\ k=0 \end{gathered}`

Then, the vertex is (-2, 0)

The y-intercept is found as follows:

`y_{\text{ int}}=y(0)=0^2+4\cdot0+4=4`

The y-intercept is (0,4)

The vertical line: x = h, in this case x = -2 is the symmetry axis. Then the point

(-4, 4) is on the parabola. With these 3 points, we can plot the function. The graph is: