1 thought on “In the xy-plane, the positive x-intercept of the graph of the equation y =x -5 is a.<br /><br />find a.<br /><br />”

Answer:

Answer

Correct option is

C

y=2(x−3)2−2

Choice A is not the correct answer. This answer displays the location of the y-value of the y-intercept of the graph (0,16) as a constant.

Choice B is not the correct answer. This answer displays the location of the y-value of the y-intercept of the graph open parentheses (0,16) as a constant.

Choice C is correct. The equation y=(2x−4)(x−4) can be written in vertex form, y=a(x−h)2+k to display the vertex,(h,k) of the parabola. To put the equation in vertex form, first multiply:

(2x−4)(x−4)

=2×2−8x−4x+16

=2×2−12x+16=2(x2−6x+8)=0

⇒x2−2×3x+9−9+8=0

⇒(x−3)2−1=0

⇒x−3=±1

∴x=3±1=4,2

y=0 for x=2,4

Choice D is not the correct answer. This answer displays the location of the x-value of one of the x-intercepts of the graph (2,0) as a constant.

Answer:Answer

Correct option is

C

y=2(x−3)2−2

Choice A is not the correct answer. This answer displays the location of the y-value of the y-intercept of the graph (0,16) as a constant.

Choice B is not the correct answer. This answer displays the location of the y-value of the y-intercept of the graph open parentheses (0,16) as a constant.

Choice C is correct. The equation y=(2x−4)(x−4) can be written in vertex form, y=a(x−h)2+k to display the vertex,(h,k) of the parabola. To put the equation in vertex form, first multiply:

(2x−4)(x−4)

=2×2−8x−4x+16

=2×2−12x+16=2(x2−6x+8)=0

⇒x2−2×3x+9−9+8=0

⇒(x−3)2−1=0

⇒x−3=±1

∴x=3±1=4,2

y=0 for x=2,4

Choice D is not the correct answer. This answer displays the location of the x-value of one of the x-intercepts of the graph (2,0) as a constant.