Question

Graph the parabolay=(x+2)^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.

Answer 1

We are asked to graph the following parabola

`y=(x+2)^2-5_{}`

Let us compare this equation with the standard vertex form.

`y=a(x-h)+k`

The vertex is the point (h, k)

So, the vertex of the given equation is (-2, -5)

Now let us find the y-intercept.

Substitute x = 0 into the given equation

`\begin{gathered} y=(0+2)^2-5_{} \\ y=(2)^2-5_{} \\ y=4-5 \\ y=-1 \end{gathered}`

So, the y-intercept point is (0, -1)

Now let us find the x-intercepts.

Substitute y = 0 into the given equation

`\begin{gathered} 0=(x+2)^2-5_{} \\ 0=x^2+2\cdot x\cdot(2)+2^2-5 \\ 0=x^2+4x+4-5 \\ 0=x^2+4x-1 \end{gathered}`

Use the quadratic formula to solve the above quadratic equation

`x=(-b\pm√(b^2-4ac))/(2a)`

Where a = 1, b = 4 and c = -1

`\begin{gathered} x=\frac{-4\pm\sqrt[]{4^2-4(1)(-1)}}{2(1)} \\ x=\frac{-4\pm\sqrt[]{16^{}+4}}{2} \\ x=\frac{-4\pm\sqrt[]{20}}{2} \\ x=\frac{-4+\sqrt[]{20}}{2},x=\frac{-4-\sqrt[]{20}}{2} \\ x=0.236,\: x=-4.236 \end{gathered}`

So, the x-intercepts are (0.236, 0) and (-4.236, 0)

Now let us plot all these points and sketch the graph of the parabola