# Find the vertex of the parabola?

how do you find the vertex

of this parabola?

y = x^2-8x+7

please and thank you

best answer will be rewarded

### 9 Answers

- cidyahLv 71 decade agoBest Answer
a=1 b=-8 c=7 if the parabola is y=ax^2+bx+c

x-coordinate of the vertex = -b / 2a

=-(-8) /2 =4

y-coordinate of the vertex is

(4)^2-8(4)+7

=16-32+7 = -9

(4,-9) is the vertex

- 1 decade ago
well since the coefficient of the x^2 is positive (in this case it's 1), then you know your parabola will be opening upward. So the vertex is on the bottom, making it a minimum. I don't know what math level you are on. But I am going to use some precal/cal.

Find the derivative of y.

y = x^2 - 8x +7

dy/dx = 2x - 8

set that equal to 0 and solve:

2x -8= 0

2x=8

x=4

plug 4 into the original equation for x.

y = 4^2 - 8(4) +7

y = 16-32+7

y = -9

vertex = (4, -9)

A simpler way is just using -b/2a to give you the x coordinate:

8/2 = 4

Then plug 4 in the original equation to get y = -9.

Source(s): I'm a math major - FrancescaLv 44 years ago
I can find the vertex of the parabola, but I am unaware of any curve named 'vertex parabola !' f(x) = x^2 - 4x + 3 = (x - 2)^2 - 1 The vertex is at: (2, - 1) QED

- TLv 51 decade ago
Take the derivative, set that equal to 0 and solve for x. Put that value into the original equation and solve for y.

dy/dx = 2x - 8 = 0

2x = 8

x = 4

y = 4^2 - 8(4) + 7

y = 16 - 32 + 7

y = -9

(4,-9)

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- 1 decade ago
This has the form ax^2+bx+x

y=1x^2-8x+7

a=1

b=-8

c=7

x intercept=7,1

y intercept=7

vertex=(4,-9)

x coordinate of the vertex=-b/2a

=-(-8)/2(1)

=8/2

=4

now plug this into the quadratic equation and find the y point

y=4^2-8(4)+7

y=16-32+7

y=-16+7

y=-9

- 6 years ago
Here a=1, b=-8andc=7

Now for any parabola ,

Vertex (x, y) is given as

x=-b/2a= 8/2=4

y=-D/4a

where D = b^2 -4ac

Therefore,

y=-[(-8)^2-(4*1*7)]/(4*1)

=-(64-28)/4=36/4=9

Hence, vertex is (4, 9)

- ComoLv 71 decade ago
Method 1

f (x) = (x ² - 8 x + 16) - 16 + 7

f (x) = (x - 4)² - 9

Vertex is V (4 , - 9)

Method 2

f `(x) = 2x - 8 = 0 for turning point

x = 4

f(4) = - 9

Vertex is V (4 , - 9)

- poornakumar bLv 71 decade ago
y = x^2-8x+7 {we will try to make a perfect square involving x}

y = x^2-8x+16 -9

y+9 = (x -4)^2

Vertex is (4, -9)