Question

How do I graph the quadratic equation $y={{x}^{2}}+4x+6$?

Answer 1

From the question, we were given to draw the graph of the quadratic equation $y={{x}^{2}}+4x+6$. Let us consider the values of x are on the x-axis and the obtained values of y on y-axis. By using the formulae for finding the discriminant of any quadratic equation $y=a{{x}^{2}}+bx+c$ given as $\Delta ={{b}^{2}}-4ac$ we can draw the graph of the given quadratic equation.

Complete step-by-step solution:
Now let us find the discriminant of the quadratic equation $y={{x}^{2}}+4x+6$.
We know that if the discriminant of quadratic equation $y=a{{x}^{2}}+bx+c$ is equal to $\Delta$, then $\Delta ={{b}^{2}}-4ac$.
So, let us assume the discriminant of the quadratic equation $y={{x}^{2}}+4x+6$is equal to $\Delta$. Then $\Delta ={{4}^{2}}-4(1)(6)=-8<0$
So, it is clear that the discriminant of the quadratic equation is less than zero.
We know that if the discriminant of the quadratic equation $y=a{{x}^{2}}+bx+c$ is less than zero, then the quadratic equation will not have roots. So, we can say that if the discriminant of the quadratic equation $y=a{{x}^{2}}+bx+c$ is less than zero, then the graph will not touch the x-axis.
So, it is clear that the graph of the quadratic equation $y={{x}^{2}}+4x+6$ does not touch the x-axis.
Now let us find whether the quadratic equation $y={{x}^{2}}+4x+6$ has local maxima or local minima.
We know that a function f(x) has local maxima or local minima at the point where the derivative of f(x) is equal to zero. So, let us find the points whether the quadratic equation $y={{x}^{2}}+4x+6$ has a local maximum or local minimum.
Let us assume the derivative of function $y={{x}^{2}}+4x+6$ is equal to ${y}'$.
${y}'=\dfrac{dy}{dx}\Rightarrow \dfrac{d}{dx}\left( {{x}^{2}}+4x+6 \right)=2x+4$
Now we should find the value of x where ${y}'$ is equal to zero.

& 2x+4=0 \\\ & \Rightarrow 2x=-4 \\\ & \Rightarrow x=-2 \\\ \end{aligned}$$ So, it is clear that at $$x=-2$$ we will have a local maxima or local minima for $$y={{x}^{2}}+4x+6$$. Now we should check whether $$x=-2$$ is a local maxima or local minima. We know that the value of x where $${f}''(x)>0$$ and $${f}'(x)=0$$ is said to be point of minima and the value of x where $${f}''(x)<0$$ and $${f}'(x)=0$$ is said to be point of maxima. So, let us assume the second derivative of $$y={{x}^{2}}+4x+6$$ is equal to $${y}''$$. $${y}''=\dfrac{d{y}'}{dx}\Rightarrow \dfrac{d}{dx}\left( 2x+4 \right)=2>0$$ So, it is clear that the second derivative of $$y={{x}^{2}}+4x+6$$ is greater than zero. So, we can say that $$x=-2$$ is the point of local minima. Now let us find the value of y where $$x=-2$$. $$\begin{aligned} & y={{\left( -2 \right)}^{2}}+4(-2)+6 \\\ & \Rightarrow y=4-8+6 \\\ & \Rightarrow y=2 \\\ \end{aligned}$$ So, we can say that the minimum value of $$y={{x}^{2}}+4x+6$$ is equal to 2 at$$x=-2$$. So, it is clear that $$\left( -2,2 \right)$$ lies on the graph $$y={{x}^{2}}+4x+6$$. So, it is clear that the graph of $$y={{x}^{2}}+4x+6$$is in the second quadrant. Now let us substitute the value of x is equal to zero in $$y={{x}^{2}}+4x+6$$. Then we get $$y=0+0+6=6$$ So, we can say that $$\left( 0,6 \right)$$ lies on the graph $$y={{x}^{2}}+4x+6$$. Now let us substitute the value of x is equal to -4. Then we get $$y={{(-4)}^{2}}+4(-4)+6=16-16+6=6$$ So, we can say that $$\left( -4,6 \right)$$ lies on the graph$$y={{x}^{2}}+4x+6$$. Now let us draw a parabola passing through $$\left( -2,2 \right)$$, $$\left( 0,6 \right)$$ and $$\left( -4,6 \right)$$ as shown below.  **Note:** We should be well aware of drawing the graphs for the quadratic equation. We should find the points by using the discriminant of the given quadratic equation. We should be very careful while plotting the points in the graph. Also, we should be very careful while finding the local minima and local maxima. This curve can be plotted without using the discriminant only by substituting $x$ values and finding the corresponding $y$ values for them.