Question

watch above video then graph this function.\\(y = (x + 3)^2 - 1\\)

Explanation:

Step1: Identify the vertex form

The function \( y = (x + 3)^2 - 1 \) is in the vertex form of a quadratic function \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. Here, \( h=-3 \) (since \( x+3=x - (-3) \)) and \( k = -1 \), so the vertex is \((-3, -1)\).

Step2: Determine the direction of opening

Since \( a = 1>0 \), the parabola opens upward.

Step3: Find the axis of symmetry

The axis of symmetry is the vertical line \( x = h \), so \( x=-3 \).

Step4: Find some additional points

• When \( x=-2 \), \( y=(-2 + 3)^2-1=1 - 1 = 0 \), so the point is \((-2, 0)\).
• When \( x=-4 \), \( y=(-4 + 3)^2-1=1 - 1 = 0 \), so the point is \((-4, 0)\).
• When \( x = 0 \), \( y=(0 + 3)^2-1=9 - 1 = 8 \), so the point is \((0, 8)\).
• When \( x=-6 \), \( y=(-6 + 3)^2-1=9 - 1 = 8 \), so the point is \((-6, 8)\).

Step5: Plot the points and draw the parabola

Plot the vertex \((-3, -1)\), the axis of symmetry \( x = -3 \), and the additional points \((-2, 0)\), \((-4, 0)\), \((0, 8)\), \((-6, 8)\). Then draw a smooth upward - opening parabola through these points.

Answer:

The graph of \( y=(x + 3)^2-1 \) is a parabola with vertex at \((-3,-1)\), axis of symmetry \( x=-3 \), opening upward, and passing through points such as \((-2,0)\), \((-4,0)\), \((0,8)\), \((-6,8)\) (the actual graph is drawn by plotting these points and drawing a smooth curve).