How do you find the range of #f(x)=-x^2+4x-3#?

Question

Kennedy Adams · Accepted Answer

Answer: # => y <= 1# Solution: The range of this function is the set of the values of #y# that the #f# can take, So, letting #y= f(x)# #y = -x^2 + 4x - 3# #=> x^2 - 4x + 3 +y = 0# For real #x# , #b^2 - 4ac >=0# For the quadratic #ax^2 + bx + c = 0# Using this in our calculation, #=> (-4)^2 - 4(1)(3 + y) >= 0# #=> 16 - 12 - 4y>= 0# #=> 4 >= 4y# #=> y <= 1# In other terms #(-oo, 1]#

Abigail Allen · Answer

undefined To find the range of $f(x) = -x^2 + 4x - 3$, determine the vertex of the quadratic function using the formula $x = -\frac{b}{2a}$. Substitute this x-value into the function to find the corresponding y-value, which represents the minimum point of the parabola. The range is then the set of all y-values greater than or equal to this minimum value.

Jameson Allen · Answer

undefined To find the range of the function f(x) = -x^2 + 4x - 3, you first need to determine its vertex, which represents the maximum or minimum point of the parabola. Then, you can analyze whether the parabola opens upwards or downwards. Based on this information, you can determine the range. The range of the function will be all real numbers if the parabola opens downwards, or it will be restricted based on the vertex if the parabola opens upwards. To find the vertex, you can use the formula for the x-coordinate of the vertex, which is given by x = -b / (2a), where a = -1 (coefficient of x^2) and b = 4 (coefficient of x). Once you find the x-coordinate of the vertex, substitute it into the function to find the corresponding y-coordinate.