Question

aph the parabola.
y=(x+1)^2+5
plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. then click on the graph-a-function button.

Explanation:

Step1: Identify the vertex form

The equation of the parabola is given in vertex form \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. For \( y=(x + 1)^2+5 \), we can rewrite it as \( y=(x - (-1))^2+5 \), so the vertex \((h, k)\) is \((-1, 5)\).

Step2: Find points to the left of the vertex

Choose \( x=-2 \) (1 unit left of \( x=-1 \)): Substitute \( x=-2 \) into \( y=(x + 1)^2+5 \), we get \( y=(-2 + 1)^2+5=(-1)^2+5 = 1 + 5=6 \). So the point is \((-2, 6)\).
Choose \( x=-3 \) (2 units left of \( x=-1 \)): Substitute \( x=-3 \) into \( y=(x + 1)^2+5 \), we get \( y=(-3 + 1)^2+5=(-2)^2+5 = 4 + 5=9 \). So the point is \((-3, 9)\).

Step3: Find points to the right of the vertex

Choose \( x=0 \) (1 unit right of \( x=-1 \)): Substitute \( x=0 \) into \( y=(x + 1)^2+5 \), we get \( y=(0 + 1)^2+5=1 + 5=6 \). So the point is \((0, 6)\).
Choose \( x=1 \) (2 units right of \( x=-1 \)): Substitute \( x=1 \) into \( y=(x + 1)^2+5 \), we get \( y=(1 + 1)^2+5=4 + 5=9 \). So the point is \((1, 9)\).

Step4: List all five points

The vertex is \((-1, 5)\), two points to the left: \((-3, 9)\), \((-2, 6)\); two points to the right: \((0, 6)\), \((1, 9)\). These points can be plotted on the graph.

Answer:

The five points on the parabola \( y=(x + 1)^2+5 \) are: Vertex \((-1, 5)\), left points \((-3, 9)\), \((-2, 6)\), right points \((0, 6)\), \((1, 9)\).