Answer with explanation:
We know that for any parabola equation of the type:
\(y=ax^2+bx+c\)
The vertex of the parabola is given by:
(h,k)
where,
\(h=\dfrac{-b}{2a}\ and\ k=\dfrac{4ac-b^2}{4a}\)
and the line of symmetry of the parabola is:
\(x=\dfrac{-b}{2a}\)
Ques 1)
We have the parabola equation as:
\(y=x^2+6x+2\)
i.e.
\(a=1\ ,\ b=6\ and\ c=2\)
Hence,
\(h=\dfrac{-6}{2\times 1}\\\\\\h=\dfrac{-6}{2}\\\\h=-3\)
and
\(k=\dfrac{4\times 1\times 2-(6)^2}{4\times 1}\\\\\\k=\dfrac{8-36}{4}\\\\k=-7\)
Hence, Vertex=(-3,-7)
and axis of symmetry is: x= -3
Ques 2)
\(y=3x^2+24x-1\)
We have:
a=3, b=24 and c=-1
Hence,
\(h=\dfrac{-24}{2\times 3}\\\\h=\dfrac{-24}{6}\\\\h=-4\)
and,
\(k=\dfrac{4\times (-1)\times 3-(24)^2}{4\times 3}\\\\\\k=-49\)
Hence, Vertex=( -4,-49)
and Axis of symmetry is: x= -4
Ques 3)
\(y=x^2+10x+25\)
\(a=1\ ,b=10\ ,c=25\)
Hence,
\(h=\dfrac{-10}{2\times 1}\\\\h=\dfrac{-10}{2}\\\\h=-5\)
and,
\(k=\dfrac{4\times (1)\times 25-(10)^2}{4\times 1}\\\\\\k=0\)
Hence, Vertex=( -5,0)
and Axis of symmetry is: x= -5
Ques 4)
\(y=-2x^2+8x-5\)
We have:
\(a=-2\ ,b=8\ and\ c=-5\)
Hence,
\(h=\dfrac{-8}{2\times (-2)}\\\\h=\dfrac{-8}{-4}\\\\h=2\)
and,
\(k=\dfrac{4\times (-2)\times (-5)-(8)^2}{4\times (-2)}\\\\\\k=3\)
Hence, Vertex=( 2,3)
and Axis of symmetry is: x= 2
Ques 5)
\(y=x^2-12x+7\)
We have:
\(a=1\ ,b=-12\ and\ c=7\)
Hence,
\(h=\dfrac{12}{2\times (1)}\\\\h=\dfrac{12}{2}\\\\h=6\)
and,
\(k=\dfrac{4\times 1\times (7)-(-12)^2}{4\times 1}\\\\\\k=-29\)
Hence, Vertex=(6,-29)
and Axis of symmetry is: x= 6
Ques 6)
\(y=x^2-4x+6\)
We have:
\(a=1\ ,b=-4\ and\ c=6\)
Hence,
\(h=\dfrac{4}{2\times (1)}\\\\h=\dfrac{4}{2}\\\\h=2\)
and,
\(k=\dfrac{4\times 1\times (6)-(-4)^2}{4\times 1}\\\\\\k=2\)
Hence, Vertex=(2,2)
and Axis of symmetry is: x= 2
