How do you find the axis of symmetry, and the maximum or minimum value of the function #y = 4x^2 + 5x – 1#?

Question

Kennedy Adams · Accepted Answer

#vertex: (-5/8, -91/16); " axis of symmetry": x = -5/8#
minimum

Given: #y = 4x^2 + 5x - 1# When the equation is in standard form: #f(x) = Ax^2 + Bx + C#, you can find the vertex and the axis of symmetry as follows: vertex: #(-B/(2A), f(-B/(2A)))#, axis of symmetry: #x = -B/(2A)# #-B/(2A) = -5/(2*4) = -5/8# #f(-5/8) = 4 (-5/8)^2 + 5(-5/8) - 1# #= 4*(25/64) - 25/8 - 1# #= 25/16 - 50/16 - 16/16 = -91/16# #vertex: (-5/8, -91/16); " axis of symmetry": x = -5/8# The vertex will be a minimum because #A# is positive.

Julia Adams · Answer

undefined To find the axis of symmetry of a quadratic function $y = ax^2 + bx + c$, use the formula $x = \frac{-b}{2a}$. For $y = 4x^2 + 5x - 1$, the axis of symmetry is $x = \frac{-5}{2(4)} = -\frac{5}{8}$. To find the maximum or minimum value, plug the value of $x$ from the axis of symmetry into the function to get the corresponding $y$ value. For this function, the maximum or minimum occurs at $x = -\frac{5}{8}$. Plug $x = -\frac{5}{8}$ into the function to find $y$. So, the maximum or minimum value is the $y$ value at $x = -\frac{5}{8}$.