Question

What is the vertex, axis of symmetry, the maximum or minimum value and the range of the parabola $y=-{{x}^{2}}+2x-5$?

Answer 1

Compare the given equation with the general form of a parabola as: - $y=a{{x}^{2}}+bx+c$. Find the respective values of a, b and c. Write the expression as: -$y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]$ and convert it into the form $\left( y+\dfrac{D}{4a} \right)=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}} \right]$. Here D = discriminant = ${{b}^{2}}-4ac$. The vertex of the parabola will be given as $\left( \dfrac{-b}{2a},\dfrac{-D}{4a} \right)$, line of symmetry will be $x=\dfrac{-b}{2a}$. Check the value of ‘a’, if it is positive then the minimum is $y=\dfrac{-D}{4a}$ and maximum is $y=\infty$. If ‘a’ is negative then the maximum value is $y=\dfrac{-D}{4a}$ and minimum value is $y=-\infty$. Write the range accordingly.

Complete step-by-step solution:
Here we have been provided with the parabolic equation: - $y=-{{x}^{2}}+2x-5$ and we are asked to find its vertex, line of symmetry, the minimum or maximum value and the range. First we need to write the vertex form of this parabola using completing the square method. Any quadratic equation of the form $y=a{{x}^{2}}+bx+c$ can be simplified as $y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]$ using completing the square method and then it can be further written as $\left( y+\dfrac{D}{4a} \right)=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}} \right]$. Here, ‘D’ denotes the discriminant. Now, on comparing $y=-{{x}^{2}}+2x-5$ with the general form we get,
$\Rightarrow$ a = -1, b = 2, c = -5
Applying the formula for discriminant of a quadratic equation given as $D={{b}^{2}}-4ac$ we get,

& \Rightarrow D={{\left( 2 \right)}^{2}}-4\left( -1 \right)\left( -5 \right) \\\ & \Rightarrow D=4-20 \\\ & \Rightarrow D=-16 \\\ \end{aligned}$$ So we can write the equation of the parabola as: - $$\begin{aligned} & \Rightarrow y=-1\left[ {{\left( x+\left( \dfrac{2}{2\times \left( -1 \right)} \right) \right)}^{2}}-\dfrac{\left( -16 \right)}{4\times {{\left( -1 \right)}^{2}}} \right] \\\ & \Rightarrow y=-1\left[ {{\left( x-1 \right)}^{2}}+4 \right] \\\ & \Rightarrow \left( y+4 \right)=-1{{\left( x-1 \right)}^{2}} \\\ \end{aligned}$$ (i) Now, the vertex of a parabola $$y=a{{x}^{2}}+bx+c$$ is a point where its maxima or minima lie depending on the case that the parabola is opening upward (positive a) or downward (negative a). The vertex is given as $$\left( \dfrac{-b}{2a},\dfrac{-D}{4a} \right)$$, so we get, $\Rightarrow $ Vertex = $\left( 1,-4 \right)$ (ii) The line of symmetry is the vertical line passing through the vertex of the parabola and it divided the parabola into two equal parts. So the equation of the line parallel to the y axis and passing through the point $\left( 1,-4 \right)$ will be given as: - $\begin{aligned} & \Rightarrow x=1 \\\ & \Rightarrow x-1=0 \\\ \end{aligned}$ (iii) Clearly we can see that the value of a is -1 and it is negative so we can conclude that the parabola will be opening downwards. In this case it will have a maximum value as $y=-4$ and minimum value as $y=-\infty $. (iv) Now, from the above minimum and maximum value we can say that the range of the function is $y\in \left[ -\infty ,-4 \right]$.  **Note:** Note that the equation $$\left( y+\dfrac{D}{4a} \right)=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}} \right]$$ is known as the vertex form of the parabola. You can also determine the minimum or maximum value of a parabolic equation with the help of differentiation. What you need to do is find $\dfrac{dy}{dx}$ and substitute it equal to 0 and find the value of x. this will be the point of maxima, substitute this value in the equation to get y. This value of y will be the maximum value of the function. These two points combined will represent the vertex of the parabola.