Yes, that is correct. An "x-

intercept" is a point on the graph where the graph crosses ("intersects") the x-axis. There y= 0. If y= (x- 2)(x+ 5)= 0 then either x- 2= 0 or x+ 5= 0. If x- 2= 0 then x= 2. If x+ 5= 0 then x= -5. In either case y= 0 so the x-intercepts are (2, 0) and (-5, 0).

No, that is

**not** correct

A. [tex]y= x^2- 9= (x- 3)(x+ 3)= 0[/tex]. We have either x- 3= 0 so x= -3 or x+ 3= 0 so x= -3. The

**two** x-intercepts are (3, 0) and (-3, 0).

B. [tex]y= x^2- 6x+ 9= (x- 3)^2= 0[/tex]. We must have x= 3. There is only one x-intercept.

C. [tex]y= x^2- 5x+ 6= (x- 3)(x- 2)= 0[/tex]. We must have either x- 3= 0 so x= 3 or x- 2= 0 so x= 2. The

**two** x-intercepts are (3, 0) and (2, 0).

D. [tex]y= x^2+ x- 6= (x+ 3)(x- 2)= 0[/tex]. We must have either x+ 3= 0 so x= -3 or x- 2= 0 so x= 2. The

**two** x-intercepts are (-3, 0) and (2, 0).

The correct answer is B.

Yes, that is correct.

A parabola opens upward if and only if its leading coefficient (the coefficient of [tex]x^2[/tex]) is positive. That is true in A and C. How "narrow" a parabola appears depends upon how large the

absolute value of the leading coefficient is. The larger the leading coefficient, the "narrower" the parabola is. |4|= 4 is larger than |-3|= 3 while |1|= 1 is not.

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