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Question: How do you find the axis of symmetry for this parabola \(y=-5{{x}^{2}}-10x-15\)?...

How do you find the axis of symmetry for this parabola y=−5x2−10x−15y=-5{{x}^{2}}-10x-15?

Explanation

Solution

Now to find the axis of symmetry we will first find the vertex of parabola. To find the vertex of parabola we will use the condition to find the extrema of the function as vertex is nothing but a minimum or maximum point on the graph. Now we will differentiate the function and equate it to zero and hence find the value of x. Now substituting the value of the x in the equation we will get the corresponding value of y. Hence we have the vertex point. Now the axis of symmetry is nothing but the line passing through the vertex and parallel to y axis.

Complete step by step solution:
Now we are given a quadratic equation y=−5x2−10x−15y=-5{{x}^{2}}-10x-15 .
To find the axis of symmetry we will first find the vertex of the graph.
Now we know that for a parabola vertex is the minimum or maximum point.
Hence we will use the condition for extremum to find the vertex.
Now we know that the condition for extremum for any graph is given by f′(x)=0f'\left( x \right)=0 .
Hence differentiating the given equation and equating it to zero we get,
⇒−10x−10=0 ⇒−10(x+1)=0 ⇒(x+1)=0 \begin{aligned} & \Rightarrow -10x-10=0 \\\ & \Rightarrow -10\left( x+1 \right)=0 \\\ & \Rightarrow \left( x+1 \right)=0 \\\ \end{aligned}
Hence the vertex is at x = - 1.
Now substituting x = - 1 in the given equation we get,
⇒y=−5(−1)2−10(−1)−15 ⇒y=−5+10−15 ⇒y=−10 \begin{aligned} & \Rightarrow y=-5{{\left( -1 \right)}^{2}}-10\left( -1 \right)-15 \\\ & \Rightarrow y=-5+10-15 \\\ & \Rightarrow y=-10 \\\ \end{aligned}
Hence the vertex of the graph represented by given equation is (−1,−10)\left( -1,-10 \right)
Now we know that the axis of symmetry is nothing but a line passing through vertex and parallel to y axis.
Now the line passing through (−1,−10)\left( -1,-10 \right) and parallel to y axis is x = - 1.

Hence equation of axis of symmetry is x = - 1.

Note: Now note that for an equation of the form ax2+bx+ca{{x}^{2}}+bx+c if we have a > 0 then the parabola is upward facing and the vertex is minimum value of the graph. Similarly if a < 0 then the parabola is downward facing and hence the vertex is the maximum point on the graph.