Question

Rewrite the following quadratic function in vertex form. Then, determine the axis of symmetry. y=5x^2+15x-2

Answer 1

5x^2+15x-2
5(x^2+3x)-2
5(x^2+3x+2.25-2.25)-2
5(x^2+3x+2.25)-13.25
5(x+1.5)^2-13.25

final answer: y=5(x+1.5)^2-13.25
axis of symmetry: (1.5, 13.25)

Answer 2

Answer:  The vertex form of the given function is $y=5\left(x+\dfrac{3}{2}\right)^2-\dfrac{53}{4}$ and its axis of symmetry is $x=-\dfrac{3}{2}.$

Step-by-step explanation:  The given quadratic function is

$y=5x^2+15x-2~~~~~~~~~~~~~~(i)$

We are to rewrite the above function in vertex form and to determine its axis of symmetry.

We have from equation (i),

$y=5x^2+15x-2\\\\\Rightarrow y=5(x^2+3x)-2\\\\\Rightarrow y=5\left(x^2+2\times x\times \dfrac{3}{2}+\dfrac{9}{4}\right)-\dfrac{45}{4}-2\\\\\Rightarrow y=5\left(x+\dfrac{3}{2}\right)^2-\dfrac{45+8}{4}\\\\\Rightarrow y=5\left(x+\dfrac{3}{2}\right)^2-\dfrac{53}{4}.$

So, the given function is a parabola with vertex at the point $\left(-\dfrac{3}{2},-\dfrac{53}{4}\right).$

Therefore, the axis of symmetry is given by

$x=-\dfrac{3}{2}.$

Thus, the vertex form of the given function is $y=5\left(x+\dfrac{3}{2}\right)^2-\dfrac{53}{4}$ and its axis of symmetry is $x=-\dfrac{3}{2}.$