For the following exercises, use the table of values that represent points on the graph of a quadratic function. By determining the vertex and axis of symmetry, find the general form of the equation of the quadratic function.

The general form of the equation of the quadratic function is x² + 2x + 1

The formula expressing the equation of the quadratic function will be expressed as:

y = a(x -h)² + k

From the table shown, we can see that the y-values are the same at where x = -2 and 0

Get the line of symmetry:

We can see that when x = -1, y = 0. Hence the vertex will be at (-1, 0)

h = -1

k = 0

Substitute into the formula y = a(x -h)² + k

y = a(x -(-1))² + 0

y = a(x+1)²

From the table at where x = 1, y = 4. On substituting to get "a" we have:

4 = a(1+1)²

4 = 4a

a = 1

Get the required quadratic function by substituting the parameters into the general formula as shown:

y = a(x -h)² + k

y = 1(x +1)² + 0

y = (x+1)²

y = (x+1) (x+1)

y = x² + x +x + 1

y = x² + 2x + 1

Hence the general form of the equation of the quadratic function is x² + 2x + 1

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y = x² + 2x + 1

Step-by-step explanation:

Same y value at x = -2 & x = 0

So line of symmetry is at:

x = (-2+0)/2

x = -1

Vertex: (-1,0)

y = a(x - -1)² + 0

y = a(x + 1)²

At x = 1, y = 4

4 = a(1 + 1)²

4 = 4a

a = 1

y = (x + 1)²

y = x² + 2x + 1

where is the figure

trapezoid

step-by-step explanation:

first i plotted all my points on a grid graph

i made sure all of them were in the correct quadrant

then i saw it was a trapezoid or a quadrilateral