How do you find the range of a quadratic equation #f(x) = -x^2 + 14x - 48#?

Question

Eva Abramson · Accepted Answer

Given #f(x) = -x^2+14x-48# #f'(x) = -2x+14#
for critical point(s)
#f'(x)= 0#
#-2x+14 = 0#
#x=7# for the only critical point We can either use our knowledge of quadratics or take the second derivative (#f''(x)=-2 rarr# slope is always decreasing) to observe that this point is a maximum. Therefore #f(x)# has a maximum when #x=7#
#f(7) = -(49) + 98 -48 = 1# So the range of #f(x)# is #[-oo,+1]#

Abigail Allen · Answer

undefined To find the range of a quadratic equation $ f(x) = -x^2 + 14x - 48 $, we first need to determine its vertex. The vertex of a quadratic function in the form $ f(x) = ax^2 + bx + c $ is given by the formula $ x = -\frac{b}{2a} $.

For the given equation $ f(x) = -x^2 + 14x - 48 $:
$$ a = -1, \quad b = 14 $$

Using the formula for the x-coordinate of the vertex:
$$ x = -\frac{b}{2a} = -\frac{14}{2(-1)} = 7 $$

Next, substitute the value of $ x $ back into the original equation to find the y-coordinate of the vertex:
$$ f(7) = -7^2 + 14(7) - 48 = -49 + 98 - 48 = 1 $$

So, the vertex of the quadratic equation is $ (7, 1) $.

Since the coefficient of $ x^2 $ is negative, the parabola opens downward. Therefore, the maximum value of the quadratic function occurs at the vertex.

Thus, the range of the quadratic equation $ f(x) = -x^2 + 14x - 48 $ is all real numbers less than or equal to the y-coordinate of the vertex, which is $ 1 $. Therefore, the range is $ f(x) \leq 1 $.