Answer: The equation of the axis of symmetry is [tex]x=-\dfrac{5}{8}[/tex] and the co-ordinates of the vertex are [tex]\left(-\dfrac{5}{8},-\dfrac{41}{16}\right).[/tex]
Step-by-step explanation: We are given to find the axis of symmetry and the coordinates of the vertex of the graph of the following function:
[tex]y=4x^2+5x-1~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
The given equation (i) describes a parabola.
The standard equation of a parabola in vertex form is given by
[tex]y=a(x-h)^2+k,[/tex]
where the co-ordinates of the vertex are (h, k) and the equation of axis of symmetry is
[tex]x=h.[/tex]
From equation (i), we have
[tex]y=4x^2+5x-1\\\\\Rightarrow y=4\left(x^2+\dfrac{5}{4}x\right)-1\\\\\\\Rightarrow y=4\left(x^2+\dfrac{5}{4}x+\dfrac{25}{64}\right)-1-\dfrac{25}{16}\\\\\\\Rightarrow y=\left(x+\dfrac{5}{8}\right)^2-\dfrac{41}{16}.[/tex]
Comparing with the vertex form, we get
[tex](h,k)=\left(-\dfrac{5}{8},-\dfrac{41}{16}\right)[/tex]
and the equation of axis of symmetry is
[tex]x=-\dfrac{5}{8}.[/tex]
Thus, the equation of the axis of symmetry is [tex]x=-\dfrac{5}{8}[/tex] and the co-ordinates of the vertex are [tex]\left(-\dfrac{5}{8},-\dfrac{41}{16}\right).[/tex]
