Question - Finding Turning Point and Line of Symmetry of Quadratic Function
Solution:
The question presented in the image is asking for the coordinates of the turning point (vertex) and the line of symmetry on the graph of the quadratic function $$ y = 2x^2 + 2x - 1 $$.To find the turning point, we first find the derivative of the function (to find the x-coordinate where the slope of the tangent is zero) and then substitute this x-coordinate back into the original function to get the y-coordinate.The derivative of $$ y = 2x^2 + 2x - 1 $$ with respect to $$ x $$ is:\[ \frac{dy}{dx} = 4x + 2 \]Setting the derivative equal to zero to find the x-coordinate of the turning point:\[ 4x + 2 = 0 \]\[ 4x = -2 \]\[ x = -\frac{1}{2} \]Substitute $$ x = -\frac{1}{2} $$ into the original equation to get the y-coordinate:\[ y = 2(-\frac{1}{2})^2 + 2(-\frac{1}{2}) - 1 \]\[ y = 2(\frac{1}{4}) - 1 - 1 \]\[ y = \frac{1}{2} - 2 \]\[ y = -\frac{3}{2} \]So the turning point's coordinates are $$ (-\frac{1}{2}, -\frac{3}{2}) $$.The line of symmetry for a parabola given by a quadratic function is always a vertical line that passes through the x-coordinate of the vertex. Thus, the line of symmetry for this graph is $$ x = -\frac{1}{2} $$.
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