Question

Given $y={{x}^{2}}-5x+4$. How do you write the equation of the axis of symmetry?

Answer 1

We know that a quadratic equation $y=a{{x}^{2}}+bx+c$ , the equation of symmetry is $y=\dfrac{-b}{2a}$. First of all, we should compare $y=a{{x}^{2}}+bx+c$ with $y={{x}^{2}}-5x+4$. From this, we have to find the values of a, b and c. From this we have to find the value of $\dfrac{-b}{2a}$. In this way, we can find the axis of symmetry.

Complete step-by-step answer:
From the question, it is given that $y={{x}^{2}}-5x+4$ and we have to find the equation of the axis of symmetry.
We know that a quadratic equation $y=a{{x}^{2}}+bx+c$ , the equation of symmetry is $y=\dfrac{-b}{2a}$.
Now we have to compare $y=a{{x}^{2}}+bx+c$ with $y={{x}^{2}}-5x+4$. Now we have to compare the both equations, we have to find the values of a, b and c respectively.
So, it is clear that the value of a, b and c are equal to 1, -5 and 4 respectively.
Let us consider

& a=1....(1) \\\ & b=-5...(2) \\\ & c=4...(3) \\\ \end{aligned}$$ We already know that a quadratic equation $$y=a{{x}^{2}}+bx+c$$ , the equation of symmetry is $$y=\dfrac{-b}{2a}$$. Now we have to find the axis of symmetry of $$y={{x}^{2}}-5x+4$$. Now we have to find the value of $$\dfrac{-b}{2a}$$. Let us assume the value of $$\dfrac{-b}{2a}$$ is equal to A. $$\Rightarrow A=\dfrac{-b}{2a}....(4)$$ Let us substitute equation (1), equation (2) and equation (3) in equation (4), then we get $$\begin{aligned} & \Rightarrow A=\dfrac{-(-5)}{2(1)} \\\ & \Rightarrow A=\dfrac{5}{2}..(5) \\\ \end{aligned}$$ So, it is clear that the equation of symmetry is $$y=\dfrac{5}{2}$$.  **Note:** Students may have a misconception that for a quadratic equation $$y=a{{x}^{2}}+bx+c$$ , the equation of symmetry is $$y=\dfrac{b}{2a}$$. But we know that a quadratic equation $$y=a{{x}^{2}}+bx+c$$ , the equation of symmetry is $$y=\dfrac{-b}{2a}$$. So, if this misconception is followed, then the final answer may get interrupted.