Find the Vertex f(x)=4x^2-8x+3

Problem

ƒ(x)=4*x2−8*x+3

Solution

1. Identify the coefficients of the quadratic function in the form ƒ(x)=a*x2+b*x+c Here, a=4 b=−8 and c=3

2. Calculate the x-coordinate of the vertex using the formula h=(−b)/(2*a)

h=(−(−8))/(2*(4))

h=8/8

h=1

3. Evaluate the function at x=1 to find the y-coordinate of the vertex, k=ƒ(h)

k=4*(1)2−8*(1)+3

k=4−8+3

k=−1

4. State the vertex as the ordered pair (h,k)

Final Answer

Vertex=(1,−1)

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