Domain and Range of a quadratic function?

Question

y = x^2 + 4x -2 and y = -x^2 + x- 7

Also I need to find the vertex, Axis of symmetry, x intercept, y intercept, increasing interval and decreasing interval. How would I go upon doing this?

Answer 1

The domain of a quadratic is all real numbers.

The range varies based on the function in question.



the vertex is at (-b/2a, f(-b/2a)). If a is positive this is the minimum, if a is negative, this is the maximum

(will tell you the range)



y=x^2+4x-2

the vertex is at (-2,-6). the axis of symmetry goes through the vertex so x=-2

the y-intercept is when x=0 so f(0)=-2, (0,-2)

the range is {y|y is an element of real numbers, y≥-6}

because it opens upward it decreases to the left of the vertex (-∞,-2)

and increase to the right of the vertex (-2,∞)

the x-intercepts are when the function = 0

you can factor, complete the square, or use the quadratic equation

x^2+4x+4-2-4=0

(x+2)^2-6=0

(x+2)^2=6

x+2=±√6

x=-2-√6 or x=-2+√6

the intercepts are (-2-√6,0),(-2+√6,0)



y=-x^2+x-7

for this one I'm going to start by completing the square because it is easier to extract most of the information (should have done this for the last one too)



y=-x^2+x-7

y=-(x^2-x+ )-7

half of -1 is -1/2, squared is 1/4. since we are adding 1/4 inside the parentheses, times -1=-1./4 we are subtracting 1/4. To keep the equation the same we need to add 1/4

y=-(x^2-x+1/4)-7+1/4



y=-(x-1/2)^2-27/4



in this form: y=a(x-h)^2+k

the vertex is (h,k) and a is the vertical stretch/direction. If a is positive it opens up, negative it opens down



the vertex is (1/2,-27/4) and it opens down so that is the maximum

the domain is all real numbers

the range is {y|y is an element of real numbers, y≤-27/4}

the axis of symmetry is x=1/2

the y intercept is (0,-7)

because this opens downward, it increases to the left of the vertex (-∞,1/2)

and decreases to the right (1/2,∞)

because the vertex is below the x-axis and opens down, there are no x-intercepts. If you solve the quadratic equation, you will get a complex solution

Answer 2

Complete the square to convert the equation to vertex form: y = a(x-h)² + k.

The vertex is (h, k).

The axis of symmetry is x = h.

The y-intercept is the value of y when x = 0.

The x-intercepts are the roots of a(x-h)²+k. If there are no real roots, there are no x-intercepts.

If a>0, the graph increases when x>h.

If a<0, the graph increases when x

Answer 3

y= x^2+4x-2

domain is (-∞ , ∞ )



y= (x^2+4x+4-4-2)

y = (x+2)^2 - 6

vertex is (-2,-6)



Since the parabola slopes upward, the range is [-6, ∞)



y=-x^2+x-7

y= -(x^2-x+7)

y= - ( x^2 - x + 1/4 - 1/4 + 7)

y = - ( (x-1/2)^2 +27/4)

y = -(x-1/2)^2 - 27/4



domain is (-∞ , ∞ )

This is a parabola with vertex (1/2, -27/4) that slopes downward

The range is (-∞, -27/4]

Answer 4

pie are square!

no, pie are round

my pie are oval or rectangular