Question

Consider the given function.

Plot the x-intercept(s), y-intercept, vertex, and axis of symmetry of the function.

Answer 1

Explanation:

To find the intercept with the x axis, we make h (x) = 0.

`0 = (x + 1) ^ 2 - 4\\0 = x ^ 2 + 2x +1 -4\\0 = x ^ 2 + 2x - 3\\0 = (x + 3)(x-1)\\x = 1\\x = -3`

To find the intercept with the y axis, we make x = 0.

`y = (0+1)^2 -4\\y = 1 - 4\\y = -3`

For a quadratic equation of the form
`ax ^ 2 + bx + c`

The vertex is calculated with the formula
`x = (-b)/(2a)`

So:

`x = (-2)/(2(1))\\\\x = -1`

Finally the axis of symmetry of a quadratic function always passes through the vertex.

So:

x = -1

Below is a graph for this function, where you can see the cut points with the axes, the vertex and axis of symmetry

Answer 2

x -intercepts are x = -3 and x = 1

y-intercept at y = -3

The line of symmetry is x = -1

vertex is (h,k) = (-1,-4)

Explanation:

Given a quadratic equations: h(x) = (x + 1)^2 - 4

The vertex for is :

h(x) = a(x - h)² + k

where,'h' is the axis of symmetry and (h,k) is the vertex.

So from the given equation we will rewrite the equation as:

h(x) = (x - (-1))^2 - 4

Hence,h = -1 and k = -4

The line of symmetry is x = -1 and vertex is (h,k) = (-1,-4)

Now, we have to find the x intercepts,

using the equation,

(x + 1)² = 4

Taking the square root on both sides,

√(x + 1)² = √4

x + 1 = ± 2

x = 1 or x = -3

So x -intercepts are x = -3 and x = 1

For y-intercept put x = 0 into the real equation:

h(x) = (0 +1)^2 - 4

y = 1 - 4

y = -3

So y-intercept at y = -3

Here with I have attached the graph.

Thank you.