Question

Consider the following quadratic function. Y=x^2+2x+7 Graph this quadratic function by identifying two points on the parabola, other than the vertex and zeros.

Answer 1

`y=x^2+2x+7`

Vertex:

`\begin{gathered} f(x)=ax^2+bx+c \\ \\ (-(b)/(2a),f(-(b)/(2a))) \end{gathered}`

For the given function:

`\begin{gathered} x-coordinate \\ -(b)/(2a)=-(2)/(2(1))=-1 \\ \\ y-coordinate \\ f(-1)=(-1)^2+2(-1)+7 \\ f(-1)=1-2+7 \\ f(-1)=6 \end{gathered}`

Then, the vertex of the given parabola is the point (-1,6)

As the leading coefficient of the parabola (a) is greather than 0 the parabola opens up. When the parabola opens up and the y-coordinate of the vertex is greather than 0 the function has not real zeros.

As the vertex is the minimum value in a parabola that opens up you need to find points that have greather values of y than the vertex. Or you can find one point on the left (x-coordinate less than -1) of the vertex and one on the right (x-coordinate greather than -1)

Point A:

When x= -3

`\begin{gathered} y=(-3)^2+2(-3)+7 \\ y=9-6+7 \\ y=10 \end{gathered}`

Point A (-3, 10)

Point B:

when x=0

`\begin{gathered} y=0^2+2(0)+7 \\ y=0+0+7 \\ y=7 \end{gathered}`

Point B: (0,7)

You get the next graph.