Question

How do you find the axis of symmetry, vertex and x intercepts for \[y={{x}^{2}}-6x+5\]?

Answer 1

Hint: Now in the given equation we will first differentiate the equation and equate to zero to find minimum of the graph. Now substituting the value of x we will find the corresponding value of y to find the coordinates of minima of the graph. This minima is nothing but vertices. Now the axis of symmetry is nothing but a line parallel to y axis passing through vertex. Hence we have the axis of symmetry. Now to find the x intercepts we will find the roots of the equation by splitting the middle term such that the product of the terms is multiplication of first and last term. Now we will simplify and find the roots.

Complete step by step solution:
Now consider the given quadratic equation \[y={{x}^{2}}-6x+5\]
Now we know that vertex is nothing but the minimum point on the graph of function.
To find vertex we will use the condition for extremum which is $f'\left( x \right)=0$
Now on differentiating the given quadratic and equating with zero we get,
$\begin{align}
  & \Rightarrow 2x-6=0 \\
 & \Rightarrow 2x=6 \\
 & \Rightarrow x=3 \\
\end{align}$
Now on substituting x = 3 in the expression we get
$\begin{align}
  & \Rightarrow y={{3}^{2}}-6\left( 3 \right)+5 \\
 & \Rightarrow y=9-18+5 \\
 & \Rightarrow y=-4 \\
\end{align}$
Hence we get the minimum point on the graph is $\left( 3,-4 \right)$
Now since this is nothing but vertex we have the coordinates of vertex is $\left( 3,-4 \right)$
Now if $\left( h,k \right)$ is the vertex of the quadratic then x = h is the axis of symmetry.
Hence we have x = 3 is the axis of symmetry for graph of \[y={{x}^{2}}-6x+5\] .
Now we want to find the x intercept for the given equation.
Now x intercept is nothing but the roots of the equation.
Now consider the equation \[{{x}^{2}}-6x+5=0\]
To find the roots we will split the middle term.
To use this method we will split the middle term of the equation such that their product is multiplication of first term and last term. Hence we have,
$\begin{align}
  & \Rightarrow {{x}^{2}}-5x-x+5=0 \\
 & \Rightarrow x\left( x-5 \right)-1\left( x-5 \right)=0 \\
 & \Rightarrow \left( x-1 \right)\left( x-5 \right)=0 \\
\end{align}$
Hence the roots of the given equation is x = 1 and x = 5.

Note: Now note that we can find the roots of quadratic by various methods. We can use the completing square method. To find roots by this method we first divide the whole equation by a a and then add or subtract ${{\left( \dfrac{b}{2a} \right)}^{2}}$ . Now we will use formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ to simplify the equation and finally take square root to find the value of x.