Find the Properties f(x)=x^2+4x-5

Problem

ƒ(x)=x2+4*x−5

Solution

1. Identify the coefficients of the quadratic function in the form ƒ(x)=a*x2+b*x+c

a=1

b=4

c=−5

2. Determine the direction of opening by looking at the sign of a Since a>0 the parabola opens upward.

3. Find the vertex (h,k) using the formula h=−b/(2*a)

h=−4/(2*(1))=−2

k=ƒ*(−2)=(−2)2+4*(−2)−5=4−8−5=−9

Vertex=(−2,−9)

4. Identify the axis of symmetry, which is the vertical line passing through the vertex x=h

Axis of Symmetry:x=−2

5. Find the y-intercept by evaluating ƒ(0)

ƒ(0)=0+4*(0)−5=−5

y-intercept=(0,−5)

6. Find the x-intercepts by setting ƒ(x)=0 and solving the quadratic equation.

x2+4*x−5=0

(x+5)*(x−1)=0

x=−5,x=1

x-intercepts=(−5,0),(1,0)

7. Determine the domain and range. The domain of any quadratic is all real numbers. Since the parabola opens upward, the range starts from the vertex ycoordinate.

Domain:(−∞,∞)

Range:[−9,∞)

Final Answer

Properties:Vertex *(−2,−9), x-intercepts *(−5,0),(1,0), y-intercept *(0,−5), Axis of Symmetry *x=−2

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