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i need help with this question for my maths classwork

I Need Help With This Question For My Maths Classwork class=

Sagot :

Note: Please note that you missed adding the option points. So, I am solving the entire function to determine the points and the vertex which would anyways clear your concept.

Answer:

The vertex is located at (3, 7)

some of the points will make the table such as:

x       0           1           2          3          4

y      -2          3           6           7          6

The graph is attached below.

Step-by-step explanation:

Given the table

[tex]y\:=\:6x-2-x^2[/tex]

The given equation is of the form quadratic function y = ax² + bx + c. Thus, it represents the Parabola.

  • As the leading coefficient of the parabola is -2, thus the Parabola opens down.

We know that the vertex would be the highest point on the Parabola graph if the graph (Parabola) opens down.

Thus, the given Parabola has the vertex point at the highest point of the Parabola as shown in the attached figure.

Finding the Vertex:

y = -x² + 6x - 2

comparing with y = ax² + bx + c

a = -1, b = 6, c = -2

Finding the x-coordinate of the Vertex

x = -b / 2a

x = [-6] /[2(-1)]

x = [-6 ] / [-2]

x = 6/2

x = 3

Thus, the x-coordinate of the vertex : x = 3

Now substituting x = 3 in the equation y = -x² + 6x - 2 to determine the y-coordinate of the vertex

y = -x² + 6x - 2

y = -(3)²+6(3)-2

y = -9 + 18 - 2

y = 18 - 11

y = 7

Thus, the y-coordinate of the vertex : y = 7

Therefore, the vertex is located at (3, 7)

Now, finding some other points lie on the Parabola

y = -x² + 6x - 2

Putting x = 0 in the main function

y = -(0)² + 6(0) - 2

y = 0 + 0 - 2

y = -2

Thus, the y-intercept lies on (0, -2)

Putting x = 1 in the main function

y = -x² + 6x - 2

y = -(1)² + 6(1) - 2

y = -1 + 6 - 2

y = 3

Thus, the point (1, 3) lies on the Parabola

Putting x = 2 in the main function

y = -x² + 6x - 2

y = -(2)² + 6(2) - 2

y = -4 + 12 - 2

y = 6

Thus, the point (2, 6) lies on the Parabola

Putting x = 3 in the main function

y = -x² + 6x - 2

y = -(3)² + 6(3) - 2

y = -9 + 18 - 2

y = 7

Thus, the point (2, 7) lies on the Parabola

Putting x = 4 in the main function

y = -x² + 6x - 2

y = -(4)² + 6(4) - 2

y = -16 + 24 - 2

y = 6

Thus, the point (4, 6) lies on the Parabola

Hence, some of the points will make the table such as:

x       0           1           2          3          4

y      -2          3           6           7          6

The graph is attached below.

View image absor201