Question - Finding Turning Point and Line of Symmetry of Quadratic Function
Solution:
The image contains a quadratic function given by $$ y = 2x^2 + 2x - 1 $$.To find the coordinates of the turning point and the line of symmetry for the graph of this parabola, we can use the vertex form of a quadratic equation or calculus methods. I'll explain both methods:**Method 1: Completing the square to find the vertex**The standard form of a quadratic function is $$ y = ax^2 + bx + c $$. To find the vertex, we can complete the square to rewrite it in vertex form $$ y = a(x - h)^2 + k $$, where (h, k) is the vertex of the parabola.For the given function $$ y = 2x^2 + 2x - 1 $$:1. Factor out the coefficient of $$ x^2 $$ from the first two terms: $$ y = 2(x^2 + x) - 1 $$2. Find the value to complete the square: $$ (\frac{1}{2})^2 = \frac{1}{4} $$3. Add and subtract this value inside the parentheses, factoring out the negative: $$ y = 2(x^2 + x + \frac{1}{4}) - 2(\frac{1}{4}) - 1 $$4. Simplify and put in vertex form: $$ y = 2(x + \frac{1}{2})^2 - \frac{1}{2} $$The vertex (turning point) of the parabola is at $$ h = -\frac{1}{2} $$, $$ k = -\frac{1}{2} $$, so the vertex is at (-0.5, -0.5).**Method 2: Using calculus to find the vertex**To find the vertex of the parabola using calculus, differentiate the function with respect to x and then find the critical point where the derivative equals zero.1. Differentiate $$ y = 2x^2 + 2x - 1 $$: $$ y' = 4x + 2 $$2. Set the derivative to zero and solve for x: $$ 0 = 4x + 2 $$ $$ x = -\frac{1}{2} $$3. Substitute the x-value back into the original function to find the y-coordinate of the vertex: $$ y = 2(-\frac{1}{2})^2 + 2(-\frac{1}{2}) - 1 $$ $$ y = 2(\frac{1}{4}) - 1 - 1 $$ $$ y = \frac{1}{2} - 2 $$ $$ y = -\frac{1}{2} $$So the vertex (turning point) is at (-0.5, -0.5).**Line of symmetry**The line of symmetry is a vertical line that passes through the x-coordinate of the vertex. For this parabola, that x-coordinate is -0.5. Therefore, the equation of the line of symmetry is $$ x = -0.5 $$.
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