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Y=-x^2+6x-5

1. What is the vertex?
2. Does it open up or down?
3. What is the y-intercept?

User Runcorn
by
8.6k points

1 Answer

2 votes

Answer:

1. Vertex: (3,4)

2. It opens down.

3. The y-intercept is -5

Explanation:

1. For a quadratic function in the form
y=ax^2+bx+c, you can calculate the x-coordinate of the vertex with:


x=(-b)/(2a)

Then, given the function
y=-x^2+6x-5, you can identify that:


b=6\\a=-1

Substituting, you get:


x=(-6)/(2(-1))=3

Substitute
x=3 into
y=-x^2+6x-5 to find the y-coordinate of the vertex:


y=-(3)^2+6(3)-5=4

Then the vertex is at (3,4)

2. You identified above the value of "a". This is:


a=-1

Then, since
a<0, the parabola opens down.

3. Substitute
x=0 into the function and solve for "y":


y=-(0)^2+6(0)-5\\y=-5

Therefore, the y-intercept is -5

User Abuzar Amin
by
8.4k points

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