Respuesta :
Funtion ! in vertex form is given by
f(x) = 4x^2 + 8x + 1 = 4(x^2 + 2x + 1/4) = 4(x^2 + 2x + 1 + 1/4 - 1) = 4(x + 1)^2 + 4(-3/4) = 4(x + 1)^2 - 3
Thus, the least minimun value is (-1, -3)
Also, the least minimum value of function 2 is (-1, 0)
Therefore, function 1 has the least minimum value at (-1, -3)
f(x) = 4x^2 + 8x + 1 = 4(x^2 + 2x + 1/4) = 4(x^2 + 2x + 1 + 1/4 - 1) = 4(x + 1)^2 + 4(-3/4) = 4(x + 1)^2 - 3
Thus, the least minimun value is (-1, -3)
Also, the least minimum value of function 2 is (-1, 0)
Therefore, function 1 has the least minimum value at (-1, -3)
Answer:Function 1 has the least minimum value and its coordinates are (-1, -3).
Step-by-step explanation:
Any quadratic function can be written in the form:
[tex]f(x)=ax^{2} +bx+c[/tex]
The x of the vertex can be fiund by x=-b÷2a
Function 1 is given by:[tex]f(x)=4x^{2} +8x+1[/tex]
Comparing the quadratic equations we have a=4,b=8,c=1.
The x of the vertex is -b÷2a=-8÷2(4)=- 8÷8= -1.
The y of the vertex can be found by substituting x= -1 in the original function and solving for y.
[tex]y=4(-1^{2} )+8(-1)+1=-3[/tex]
Vertex of f(x)=(-1,-3)
For function g(x) minimum is at (0,0)
Hence Function 1 has the least minimum value and its coordinates are (-1, -3).