a math question?

Question

find the co-ordinates of the vertex and the axis of symmetry:
f(x)=x^2+4x-7

thank you

Answer 1

Hi,

The axis of symmetry is found by x = -b/(2a). In this case,

x = -4/(2*1) = -2

The axis of symmetry is x = -2.

To find the vertex, substitute -2 for x in the equation.

f(-2)=(-2)^2 + 4(-2) - 7
f(-2) = 4 - 8 - 7 = -11

The vertex is at (-2,-11)

I hope that helps!! :-)

Answer 2

To find the axis of symmetry use the equation -b/(2a) = x
f(x) = x^2 + 4x - 7 where f(x) = ax^2 +bx + c
therefore a = 1, b = 4, c = -7
so x = -b/2a = -4/[(2)(1)] = -4/2 = -2

Therefore the axis of symmetry is x = -2.

It can also be found by equating the derivative of f(x), f'(x) to 0
the derivative f'(x) = 2x + 4
2x + 4 = 0
2x=-4
x=-2

To find the coordinates of the vertex, substitute x= -2 into the equation to find the corresponding y value.
sub x = -2:
f(-2)=(-2)^2 + 4(-2) - 7
f(-2) = 4 - 8 - 7 = -11

The vertex is therefore at (-2,-11)

Hope this helps =]

Answer 3

Since this is the equation of a parabola that opens 'upwards' (since the x² term is positive) the vertex will be at the point the derivative of the fuinction is 0
dy/dx = 2x + 4 = 0 => x = -2 and
(-2)² + 4(-2) - 7 = -11so the vertex is at (-2, -11) and the axis of symmetry is x = -2

Doug

Answer 4

Complete the squre first:
x^2 + 4x - 7 = x^2 + 4x + 2^2 - 2^2 -7
= (x + 2)^2 - 4 - 7 = (x + 2)^2 -11

Equate the expression in the parentheses to 0:
x + 2 = 0
x = -2

It gives the turning point of the curve. Also, it gives the axis of symmetry. The curve is symmetric about x = -2

The value outside the square gives the maximum or minimum value of the function (in this case, MINIMUM)

Therefore, the MINIMUM value of the function = -11

Note:
If the co-efficient of the square term is positive then the value outside the square gives the MINIMUM point curve.

If the co-efficient of the square term in negative the value outside the square gives the MAXIMUM point of the curve.

I hope this helps.

Answer 5

The above equation is equation of curve
By Solving
x^2 + 4x - 7 = x^2 + 4x + 2^2 - 2^2 -7
= (x + 2)^2 - 4 - 7 = (x + 2)^2 -11

Equate the expression in the parentheses to 0:
x + 2 = 0
x = -2

It gives the turning point of the curve.

The value outside the square gives the maximum or minimum value of the function (in this case, MINIMUM)

Therefore, the MINIMUM value of the function = -11


Complete the squre first:
x^2 + 4x - 7 = x^2 + 4x + 2^2 - 2^2 -7
= (x + 2)^2 - 4 - 7 = (x + 2)^2 -11

Equate the expression in the parentheses to 0:
x + 2 = 0
x = -2

It gives the turning point of the curve.

The value outside the square gives the maximum or minimum value of the function (in this case, MINIMUM)

Therefore, the MINIMUM value of the function = -11

Note:
If the co-efficient of the square term is positive then the value outside the square gives the MINIMUM point curve.

If the co-efficient of the square term in negative the value outside the square gives the MAXIMUM point of the curve

Answer 6

f(x)=x^2+4x-7
f(x)=ax^2+bx+c
the axis of symmetry=-b/2a
=-4/2(1)
=-2
the axis of symmetry:x=-2


the x co-ordinates of the vertex =the axis of symmetry
=-2

the y co-ordinates of the vertex ,
f(-2)=(-2)^2+4(-2)-7
=4-8-7
=-11

the co-ordinates of the vertex:(-2,-11)

Answer 7

Method 1
f `(x) = 2x + 4
f "(x) = 2 (gives MIN. turning point)
2x + 4 = 0 for turning point
x = - 2
f(- 2) = 4 - 8 - 7
f(-2) = - 11
Vertex (- 2 , -11)
Axis of symmetry is x = - 2

Method 2
y = (x² + 4x + 4) - 4 - 7
y = (x + 2)² - 11
Vertex (- 2, - 11)
Axis of symmetry is x = - 2