Example Question - vertex form
Here are examples of questions we've helped users solve.
Finding the Equation of a Parabola with Specific Properties
The question in the image is asking for an equation of a parabola that opens upwards, has a minimum value of 8, and has an axis of symmetry at \(x = 3\). To find the correct equation, let's recall some properties of parabolas: 1. A parabola that opens upwards will have the form \(f(x) = a(x - h)^2 + k\), where \(a > 0\). 2. The vertex of the parabola is at the point \((h, k)\). This will also be the minimum point if the parabola opens upwards. 3. The axis of symmetry is the vertical line \(x = h\). Based on the given information, we need the parabola to open upwards which requires that \(a > 0\), its vertex will be at \((3, 8)\), meaning \(h = 3\) and \(k = 8\), and so the axis of symmetry is \(x = 3\). All the options in the image are in the vertex form of a parabola, \(f(x) = a(x - h)^2 + k\). We need to find the one that matches our vertex \((3, 8)\) and opens upwards. - Option A has \(h = 3\) and \(k = 8\) which is correct, and since \(a = 1\) (which is positive), this parabola opens upwards. This equation could be the correct answer. - Option B has the same \(h\) and \(k\), but a negative \(a\), meaning this parabola opens downwards. This cannot be the correct answer. - Option C has the correct \(k\), but \(h = -3\), which gives us the wrong axis of symmetry. So it's not correct. - Option D has \(h = 3\) and \(k = -8\), which means the vertex is at \((3, -8)\) and this does not match our minimum value of 8. Therefore, the correct answer is Option A: \(f(x) = (x - 3)^2 + 8\), because this is the only equation among the choices that correctly represents a parabola with the specified properties.
Identifying a Parabola with Specific Characteristics
The question in the image asks which equation represents a parabola that opens upward, has a minimum value of 3, and has an axis of symmetry at \( x = -3 \). Parabolas that open upward have a positive coefficient in front of the \( x^2 \) term. A minimum value is represented by the vertex of the parabola, and in the vertex form of a parabola, \( y = a(x - h)^2 + k \), where the vertex is at the point \( (h, k) \), \( k \) will be the minimum value when the parabola opens upward. The axis of symmetry is at \( x = h \). Looking at the options given: A. \( f(x) = (x - (-3))^2 + 3 \) B. \( f(x) = -(x - (-3))^2 + 6 \) C. \( f(x) = (x - 3)^2 + 6 \) D. \( f(x) = (x - 3)^2 + 3 \) Option A, \( f(x) = (x - (-3))^2 + 3 \), simplifies to \( f(x) = (x + 3)^2 + 3 \), which has the correct axis of symmetry at \( x = -3 \) and a minimum value of 3. The coefficient in front of \( (x + 3)^2 \) is positive, indicating that the parabola opens upward. Options B, C, and D either do not have the correct axis of symmetry, or they have a negative leading coefficient (which would mean the parabola opens downward), or they don't have the correct minimum value. Therefore, the correct option is A. \( f(x) = (x + 3)^2 + 3 \).
Identifying a Parabolic Equation with Given Characteristics
The image contains a question asking which equation represents a parabola that opens upward, has a minimum value of 3, and has an axis of symmetry at \( x = -3 \). The options provided are four quadratic equations: A. \( f(x) = -(x + 3)^2 + 3 \) B. \( f(x) = (x - 3)^2 + 6 \) C. \( f(x) = (x + 3)^2 - 6 \) D. \( f(x) = (x + 3)^2 + 3 \) A parabola that opens upward will have the square term with a positive coefficient. Options B, C, and D all have the square term with a positive coefficient, and thus they represent parabolas that open upward. Option A is incorrect because the negative sign in front would make the parabola open downward. Regarding the axis of symmetry, it is determined by the \( h \) in the vertex form of a parabola \( f(x) = a(x - h)^2 + k \), where \( (h, k) \) is the vertex of the parabola. The axis of symmetry is x = h. So in this case, we need \( h = -3 \). Options C and D both have \( x + 3 \), which can be written as \( x - (-3) \), indicating that the axis of symmetry is at \( x = -3 \), which meets the axis of symmetry criteria. Lastly, the parabola has a minimum value of 3, which means that the vertex is at \( (h, k) = (-3, 3) \). So we want the constant term (after completing the square) to be 3. Only option D has a constant term of +3. The correct answer based on all these conditions is: D. \( f(x) = (x + 3)^2 + 3 \) This equation represents a parabola that opens upward, has its vertex and therefore its minimum value at 3, and has an axis of symmetry at \( x = -3 \).
Quadratic Function Equation Calculation from Graph
To find the equation of the quadratic function g whose graph is shown, we will use the vertex form of a quadratic function, which is: \[ g(x) = a(x - h)^2 + k \] where (h, k) is the vertex of the parabola, and 'a' is a coefficient that determines the width and direction (upward or downward) of the parabola. From the graph, we can see that the vertex of the parabola is at (-4, 2). Therefore, h = -4 and k = 2. To find 'a', we need another point on the graph. We can use the point (-5, 0), which is one of the x-intercepts. Now, substitute h, k, and the coordinates of the point (-5, 0) into the vertex form and solve for a: \[ 0 = a(-5 - (-4))^2 + 2 \] \[ 0 = a(-5 + 4)^2 + 2 \] \[ 0 = a(1)^2 + 2 \] \[ 0 = a + 2 \] \[ a = -2 \] So the coefficient 'a' is -2. Now we can write the complete equation of the function: \[ g(x) = -2(x - (-4))^2 + 2 \] \[ g(x) = -2(x + 4)^2 + 2 \] That is the equation of the quadratic function g whose graph is shown in the image.
Solving Quadratic Equations and Finding Equation in Standard Form
To find the equation of a quadratic function in the form of f(x) = ax^2 + bx + c, we should identify key points on the graph to help us determine the values of a, b, and c. From the graph provided, there are two points we can use: 1. The vertex of the parabola, which is (3, -9). 2. Another point on the curve, which is (6, 6). The vertex form of a quadratic equation is given by: f(x) = a(x - h)^2 + k, where (h, k) is the vertex of the parabola. Since we know the vertex is (3, -9), the vertex form of our equation becomes: f(x) = a(x - 3)^2 - 9. We can use the other given point (6, 6) to find the value of 'a'. Plugging the point into the vertex form gives us: 6 = a(6 - 3)^2 - 9 6 = a(3)^2 - 9 6 = 9a - 9 Now, add 9 to both sides to get: 6 + 9 = 9a 15 = 9a Divide both sides by 9 to solve for 'a': 15 / 9 = a a = 5/3. Now we can write the complete equation for the parabola: f(x) = (5/3)(x - 3)^2 - 9. To write it in standard form, we can expand the squared term and distribute 'a': f(x) = (5/3)(x^2 - 6x + 9) - 9 f(x) = (5/3)x^2 - 10x + 15 - 9 f(x) = (5/3)x^2 - 10x + 6. So the equation of the quadratic function in standard form is: f(x) = (5/3)x^2 - 10x + 6.
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 \( x = -0.5 \).
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