Y Intercept

[SWood]

In one of my assignments I was asked to find the x and y intercepts of y=(-1/2)x^2+2x+10

To find the x intercept, I let y=0
0=(-1/2)x^2+2x+10
I solved the equation with the quadraitic formula
I got x=-2.9 or6.9
This is correct.


To find the y intercept, I let x=0
y=(-1/2)(0)^2+2(0)+10
y=0
y intercept is (0,10)
I got this assignment back saying "The author has correctly determined the x value of the vertex, however, the y value is not correct."

 

[wjm11]

y=(-1/2)x^2+2x+10

To find the y intercept, I let x=0
y=(-1/2)(0)^2+2(0)+10
y=0
y intercept is (0,10)
I got this assignment back saying "The author has correctly determined the x value of the vertex, however, the y value is not correct."

Your work is fine, though you have a typo where you wrote y = 0. You correctly stated, however, that the y-intercept is (0,10).

I cannot explain why your answer was deemed incorrect (by a software application?). You might try submitting it again. Make sure you didn't accidentally use the letter O in place of the number 0, or something like that. Software can be rather picky about those things.

 

[SWood]

It asks for x intercepts, y intercepts and the vertex.

I got x intercepts at (-2.9,0) and (6.9,0)
y intercept at (0,10)
and Vertex at (2,12)

Is that correct?

 

[masters]

It asks for x intercepts, y intercepts and the vertex.

I got x intercepts at (-2.9,0) and (6.9,0)
y intercept at (0,10)
and Vertex at (2,12)

Is that correct?

Hi SWood,

Good job. Everything looks good.