Let () = βˆ’22 βˆ’ 8 + 4. Then (a) write the function in vertex form. (b) identify the vertex. (c) determine the x-intercept(s). (d) determine the y-intercept. (e) sketch the function. (f)
determine the axis of symmetry. (g) determine the minimum or maximum value of the function. (h) write the domain and range

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Answer:

Graph the parabola y=x2βˆ’7x+2 .

Compare the equation with y=ax2+bx+c to find the values of a , b , and c .

Here, a=1,b=βˆ’7 and c=2 .

Use the values of the coefficients to write the equation of axis of symmetry .

The graph of a quadratic equation in the form y=ax2+bx+c has as its axis of symmetry the line x=βˆ’b2a . So, the equation of the axis of symmetry of the given parabola is x=βˆ’(βˆ’7)2(1) or x=72 .

Substitute x=72 in the equation to find the y -coordinate of the vertex.

y=(72)2βˆ’7(72)+2    =494βˆ’492+2    =49β€‰βˆ’β€‰98 + 84     =βˆ’414

Therefore, the coordinates of the vertex are (72,βˆ’414) .

Now, substitute a few more x -values in the equation to get the corresponding y -values.

x y=x2βˆ’7x+2

0 2

1 βˆ’4

2 βˆ’8

3 βˆ’10

5 βˆ’8

7 2

Plot the points and join them to get the parabola.

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