# Vertex of the parabola?

Find the x-intercept(s) and the coordinates of the vertex for the parabola y=x^2+10x+21. Please help me find the x intercept and the vertex

### 6 Answers

- intc_escapeeLv 71 decade agoBest Answer
The vertex form of a vertical parabola's equation is generally expressed as: y = a(x - h)² + k where (h,k) is the vertex.

y = x² + 10x + 21

y = (x + 5)² - 4 .... in vertex form: h = -5, k = -4

(x + 5)² = 4 ......... set y=0 to find x intercepts

x + 5 = ± 2

x = -3, -7

Answer: vertex: (-5,-4) ... x-intercepts: (-3,0) and (-7,0)

- 1 decade ago
The x intercept is when you set y = 0, because when y = 0, the x points that you find are the x-intercepts (do it visually if you're confused). Therefore,

x^2+10x+21 = 0

(x + 3)(x + 7) = 0

x = -3 or x = -7

The x coordinate for the vertex is given by the equation

-b/2a. To find b and a, you have to refer to the standard form for a quadratic equation, which is ax^2 + bx + c. As you can see from the given equation, a = 1 and b = 10

x^2+10x+21

Therefore, -b/2a = -5. But that's just the x coordinate. To get the y coordinate, substitute -5 in for x in the equation. So you get 25 - 50 + 21 which is -4. Therefore, the vertex is

(-5, -4)

- CatherineLv 44 years ago
A parabola of the form y-k = (x-h)^2 or y=(x-h)^2+k has (h,k) as the vertex. In all of your problems, the x-coordinate of the vertex is 0 as x^2=(x-0)^2 y=x^2 vertex(0,0) y=x^2+5 vertex (0,5) y=x^2-3 vertex (0,-3) y=x^2-2 vertex (0,-2) To complete the square x^2+4x, divide the coefficient of x by 2. (x+2)^2=x^2+4x+4 so, x^2+4x = (x+2)^2-4 (x^2-12x) same principle. x^2-12x=(x-6)^2-36

- 1 decade ago
to find the vertex you first have to find the axis of symmetry

y=x^2+10x+21 ----> use the equation x= -b/ 2a

x= -10/2(1) = -5 ----> that -5 is the axis of symmetry and also the x in the vertex.

then you put the -5 in for the x's in the equation.

y= -5^2 + 10(-5) + 21 ----> 25-50+21= 6

vertex = (-5, 6)

hope this helps!

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- 1 decade ago
well to find x intercepts you jsut set y equal to 0

so x^2+10x+21=0

then you can jsut factor that into (x+3)(x+7) and the answers are -3,-7

to find the vertex you can differentiate the fucntion and set that to 0

dy/dx=2x+10

0=2x+10

2x= -10

x= -5

or if oyu havnt had calc you remember the formula that x=-b/2a

so x=-10/2= -5 and then jsut polug that x in to find y

- 1 decade ago
I guess i will just give you the answers...

x= -3

x= -7

vertex= (-5,-4)

Those should be correct

Source(s): I just did the quadratic formula and checked with calculator