a)Find axis of symmetry b) Find turning point of parabola...of y=x^2-6x+4.?

Question

Q2. Create a quaratic equation that have roots of 1/2 and -2.

Answer 1

A)
ok a quadratic function is of the form y=ax^2+bx+c


y=x^2-6x+4
a=1
b=-6
c=4
the parabola opens up if a is positive, and opens down if a is negative
this parabola opens up
the formula for the vertex is -b/(2a)
so vertex is -(-6)/(2*1) =6/2 = 3
so the vertex is 3 so the coordinates of the vertex is
y=3^2-6*3+4=9-18+4=-9+4 = -5

so the vertex is (3,-5)
now the vertex was 3 for the axis of symmetry we use x=-b/(2a)

so the axis of symmetry is along x=3

so the turning point of any parabola is the vertex/axis


b) Create a quadratic equation that have roots of 1/2 and -2.
ok the roots of the quadratic equation give x values, the roots are where the parabola crosses the x axis so the value of y is equal to zero. trhe vertex will be halfway between the roots of the parabola
so the roots are 1/2, -2
so the value in between is -3/4 so a value that would work is 3 for b and 2 for a etc

so we need a c so lets put in a=2 and b=3
0=2(-2)^2+3(-2)+c
0=8-6+c
0=2+c
c= -2

so a suitable quadratic equation is y=2x^2+3x-2

Answer 2

Given the parabola y = x² - 6x + 4

b) Find the turning point.
The turning point is also known as the vertex. Let's put the equation into vertex form by completing the square.

y = x² - 6x + 4
y = (x² - 6x + 9) - 5
y - 5 = (x - 3)²

The vertex or turning point is (3,5).

a) Find axis of symmetry
Since the squared term is x the parabola is vertical so the axis of symmetry is vertical also and runs thru the vertex.

The axis of symmetry is:
x = 3
___________________

2) Create a quadratic equation with roots {1/2,-2}.

y = (2x - 1)(x + 2) = 2x² + 3x - 2

Answer 3

A. There is no axis of symmetry.
B. The turning point (vertex) is (3, 5).
c. 2x^2-3x-2

Answer 4

If you used a ghraping calc it'd be a whole lot easier