Example Question - line of symmetry
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Finding Turning Point and Line of Symmetry of Quadratic Function
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 \( x = -0.5 \).
Finding Turning Point and Line of Symmetry of Quadratic Function
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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