Question

What is the axis of symmetry and vertex for the graph $y = 2{\left( {x + 3} \right)^2} + 6$ ?

Answer 1

Hint : Here in this question, we have to write the axis of symmetry and vertex for the graph of the given equation. The given equation resembles or recognised as the standard equation for a parabola in vertex form, which is $f\left( x \right) = a{\left( {x - h} \right)^2} + k$ , where $\left( {h,k} \right)$ is the vertex of the parabola and next find the axis of symmetry by using a formula $x = h$ or $x = - \dfrac{b}{{2a}}$ . On simplification to get the required solution.

Complete step by step solution:
We know, in the quadratic equation $f\left( x \right) = a{x^2} + bx + c$ , a b and c are the constants and $x$ is the variable. So, by finding the different values of $x$ and corresponding values of $y$ or $f\left( x \right)$ , we can plot all the points in the graph and by joining all of them we can get the required shape.
here Parabola Formula for the equation of a parabola given in its vertex form is: $f\left( x \right) = a{\left( {x - h} \right)^2} + k$ where $\left( {h,k} \right)$ is the coordinates of the vertex.
Axis of symmetry is a line that divides an object into two equal halves, thereby creating a mirror-like reflection of either side of the object. The word symmetry implies balance.
For a parabola, the axis of symmetry is given by the formula $x = h$ for the equation $f\left( x \right) = a{\left( {x - h} \right)^2} + k$ and $x = - \dfrac{b}{{2a}}$ for quadratic equation $f\left( x \right) = a{x^2} + bx + c$ .
Where, a and b are coefficients of ${x^2}$ and $x$ respectively and c is a constant term.
Now Consider, the given equation
$\Rightarrow y = 2{\left( {x + 3} \right)^2} + 6$ ------(1)
This equation resembles as a equation of vertex form of parabola $f\left( x \right) = a{\left( {x - h} \right)^2} + k$
Here, $h = - 3$ and $k = 6$
So, vertex is
$\Rightarrow \left( {h,k} \right) = \left( { - 3,6} \right)$ .
Axis of symmetry $x = h$
$\Rightarrow x = - 3$
Axis of symmetry also find by
On expanding ${\left( {x + 3} \right)^2}$ in equation (1) using identities ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ , then
$\Rightarrow y = 2\left( {{x^2} + {3^2} + 2\left( x \right)\left( 3 \right)} \right) + 6$
$\Rightarrow y = 2\left( {{x^2} + 9 + 6x} \right) + 6$
$\Rightarrow y = 2{x^2} + 18 + 12x + 6$
On simplification, we get
$\Rightarrow y = 2{x^2} + 12x + 24$
Where, $a = 2$ , $b = 12$ and $c = 24$
Then axis of symmetry is $x = - \dfrac{b}{{2a}}$ , on substituting, we have
$\Rightarrow x = - \dfrac{{12}}{{2\left( 2 \right)}}$
$\Rightarrow x = - \dfrac{{12}}{4}$
On simplification, we get
$\Rightarrow x = - 3$
Therefore, the axis of symmetry passes through the vertex, is vertical with equation $x = - 3$ .
Hence, for the graph $y = 2{\left( {x + 3} \right)^2} + 6$ , vertex is at $\left( { - 3,6} \right)$ and axis of symmetry is $x = - 3$ .

Note : When solving these type of questions, we have to know the standard equation or general form equation of the curves like hyperbola, parabola, ellipse, circle etc each curving having a different axis of symmetry, latus rectum, focus and vertex formula so we remember that.