Example Question - vertex of parabola
Here are examples of questions we've helped users solve.
Calculating Time to Reach Maximum Height of Ball
To find out how long it takes for the ball to reach its maximum height, we need to analyze the given quadratic equation for height \( h \), which is: \[ h(t) = -16t^2 + 32t + 5 \] This is a quadratic equation in the standard form: \[ h(t) = at^2 + bt + c \] In this equation, \( h(t) \) represents the height of the ball at any time \( t \), measured in seconds. The ball will reach maximum height at the vertex of the parabola, which is given by the formula \( t = -\frac{b}{2a} \), where \( a \) and \( b \) are coefficients from the quadratic equation. Here, \( a = -16 \) and \( b = 32 \), so let's calculate: \[ t = -\frac{b}{2a} = -\frac{32}{2(-16)} = -\frac{32}{-32} = 1 \] Therefore, the ball will reach its maximum height after 1 second.
Interpreting a Quadratic Graph
The image displays a graph with a curve, but unfortunately, the question you're asking for help with is not visible in the image. I can, however, provide some general observations about the graph: - The curve appears to be a plot of a quadratic function, which is typically expressed in the form \( y = ax^2 + bx + c \), where \(a\), \(b\), and \(c\) are constants. - The graph indicates that the quadratic function has one real root (zero) where the curve crosses the x-axis. - The vertex of the parabola (the highest or lowest point on the curve) is located in the second quadrant of the Cartesian coordinate system, implying the "a" coefficient in the quadratic equation is negative, as the parabola opens downward. - The y-intercept (the point where the curve crosses the y-axis) is above the origin, suggesting that the "c" term in the equation is positive. To provide a more specific solution or analysis, I would need to know the actual question or the function that was used to generate this curve. If you wanted to find the equation of the curve, you would require either its vertex and one other point or the y-intercept and two other points on the curve. If the question is to analyze characteristics such as intercepts, vertex, axis of symmetry, maxima/minima, or intervals of increase/decrease, please provide those details.
Identifying a Parabolic Function with Specific Characteristics
The question in the image is asking which equation represents a parabola that opens upwards, has a minimum value of 3, and has an axis of symmetry at \( x = 3 \). For a parabola in the form \( f(x) = a(x-h)^2 + k \), if \( a > 0 \), it opens upwards. The vertex of the parabola is at the point \( (h, k) \), where \( h \) is the x-coordinate of the vertex and gives the axis of symmetry, and \( k \) is the y-coordinate of the vertex, representing the minimum or maximum value of the function, depending on whether the parabola opens upwards or downwards. Given that the parabola has a minimum value of 3, \( k = 3 \), and the axis of symmetry is at \( x = 3 \), so \( h = 3 \). The value of \( a \) merely determines the "width" of the parabola but is always positive for an upward-opening parabola. Considering these factors, the correct equation must have a vertex at \( (3, 3) \) and have a positive \( a \) value. This leads us to an equation of the form \( f(x) = a(x-3)^2 + 3 \). Looking at the options in the image, the only equation that matches this form is: Option D: \( f(x) = a(x-3)^2 + 3 \) The precise value of \( a \) is not given, but it is not necessary in this case because we are only concerned with the direction in which the parabola opens, the minimum value, and the axis of symmetry. All these are satisfied by Option D.
Quadratic Function Equation Determination
The graph in the image depicts a quadratic function, which can be represented by the general equation: \[ g(x) = a(x - h)^2 + k \] where \((h, k)\) is the vertex of the parabola, and \(a\) is a coefficient that determines the direction and width of the parabola. If \(a\) is positive, the parabola opens upwards; if \(a\) is negative, it opens downwards. In this graph, the vertex of the parabola is clearly at the point \((2, 7)\). Therefore, \(h = 2\) and \(k = 7\). Now, we can substitute these values into the equation: \[ g(x) = a(x - 2)^2 + 7 \] We have another point on the graph, \((3, 4)\). We can use this point to solve for the coefficient \(a\). Substituting \(x = 3\) and \(g(x) = 4\) into the equation above, we get: \[ 4 = a(3 - 2)^2 + 7 \] \[ 4 = a(1)^2 + 7 \] \[ 4 = a + 7 \] \[ a = 4 - 7 \] \[ a = -3 \] Now that we have the value of \(a\), we can write the final equation of the quadratic function: \[ g(x) = -3(x - 2)^2 + 7 \] This is the required equation of the quadratic function.
Graphing a Quadratic Function
La imagen muestra una expresión matemática y la instrucción "Dibujo gráfica", lo que sugiere que se debe dibujar la gráfica de la función dada. La expresión es una función cuadrática de la forma \( y = -x^2 + 8x - 5 \). Para dibujar la gráfica de esta función, normalmente seguiríamos estos pasos: 1. Encontrar el vértice de la parábola. 2. Determinar si la parábola se abre hacia arriba o hacia abajo. 3. Hallar los puntos de intersección con los ejes \(x\) y \(y\), si existen. 4. Dibujar la parábola utilizando los puntos encontrados para guiar el trazo. Vamos a resolver cada paso: 1. El vértice de una parábola en forma estándar \( y = ax^2 + bx + c \) se puede encontrar con la fórmula \( x = -\frac{b}{2a} \), y luego reemplazando ese valor de \(x\) en la ecuación para encontrar \(y\). Para nuestra función, \( a = -1 \) y \( b = 8 \), por lo que: \( x_v = -\frac{8}{2(-1)} = 4 \) Al reemplazar \( x = 4 \) en la ecuación, encontramos \( y \): \( y_v = -(4)^2 + 8(4) - 5 \) \( y_v = -16 + 32 - 5 \) \( y_v = 11 \) Por lo tanto, el vértice de la parábola es \( (4, 11) \). 2. Como el coeficiente \( a \) es negativo, \( a = -1 \), la parábola se abre hacia abajo. 3. Para encontrar los puntos de intersección con el eje \( x \) (las raíces de la ecuación), igualamos \( y \) a cero y resolvemos para \( x \): \( 0 = -x^2 + 8x - 5 \) Esta es una ecuación cuadrática que podemos resolver por factorización, completando el cuadrado o utilizando la fórmula cuadrática. No es fácilmente factorizable, así que vamos a usar la fórmula cuadrática: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) Donde \( a = -1 \), \( b = 8 \), y \( c = -5 \). \( x = \frac{-8 \pm \sqrt{8^2 - 4(-1)(-5)}}{2(-1)} \) \( x = \frac{-8 \pm \sqrt{64 - 20}}{-2} \) \( x = \frac{-8 \pm \sqrt{44}}{-2} \) \( x \) tendrá dos valores, uno para \( \pm \), pero como estamos trabajando con números irracionales, puedes dejarlos como parte de la expresión o encontrar valores aproximados. Por último, el punto de intersección con el eje \( y \) es simplemente \( y = -5 \) cuando \( x = 0 \). 4. Podemos esbozar la parábola a partir de estos puntos y sabiendo que se abre hacia abajo con vértice en \( (4, 11) \). Al dibujar la gráfica, debería parecer una "U" invertida, cruzando el eje \( y \) en \( -5 \) y teniendo un vértice alto en \( (4, 11) \). La parábola cruzará el eje \( x \) en los puntos encontrados con la fórmula cuadrática (en este caso se necesitaría una calculadora para obtener valores aproximados).
Determining the Range of a Function Graph
To find the range of the function graphed on the coordinate plane, you want to determine the set of all possible output (y) values that the function can produce. By observing the graph, you can see that the function reaches a highest point (the vertex of the parabola) at y = 5 and it does not go above this value. The graph continues indefinitely towards the x-axis but doesn't actually reach it, meaning it approaches but never touches or crosses the x-axis, suggesting that the range includes all y-values less than 5 but not including 5. Therefore, the range of the function is: y < 5 This can also be represented using interval notation as: (-∞, 5)
Quadratic Function Maximum and Range
The given function is \( y = -x^2 + 4x + 8 \). This is a quadratic equation in standard form \( y = ax^2 + bx + c \) with a = -1, b = 4, and c = 8. Since the coefficient 'a' is negative (-1), the parabola opens downwards which means the vertex represents the maximum point on the graph of the function. The x-coordinate of the vertex can be found by using the formula \( x = -\frac{b}{2a} \). For the given quadratic function, \( x = -\frac{4}{2 \cdot (-1)} = -\frac{4}{-2} = 2 \). Now, to find the maximum y-value (which will be the range's upper limit since the parabola opens downwards), we substitute x = 2 back into the original equation. \( y = -2^2 + 4 \cdot 2 + 8 \) \( y = -4 + 8 + 8 \) \( y = 4 + 8 \) \( y = 12 \) Therefore, the vertex of the parabola is at (2, 12), and since the parabola opens downwards, the range of y is all real numbers that are less than or equal to 12. The range is \( y \leq 12 \). The correct answer is B. \( y \leq 12 \).
Analyzing Quadratic Functions and Real Roots
Certainly! To solve the problem given in the image, we need to find the intercepts and the coordinates of the turning point of the parabola given by the quadratic function \( y = -2x^2 + 4x + 3 \), and then explain why \( y = -2x^2 + 4x - 3 \) has two distinct real roots using the graph. **Finding the Intercepts**: 1. The **y-intercept** is found by setting \( x = 0 \) in the equation. 2. The **x-intercepts** (or roots) are found by setting \( y = 0 \) and solving the quadratic equation. **Y-intercept**: \( y = -2(0)^2 + 4(0) + 3 \) \( y = 3 \) The y-intercept is at (0, 3). **X-intercepts**: \( 0 = -2x^2 + 4x + 3 \) To solve this, we can use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -2 \), \( b = 4 \), and \( c = 3 \). \( x = \frac{-4 \pm \sqrt{4^2 - 4(-2)(3)}}{2(-2)} \) \( x = \frac{-4 \pm \sqrt{16 + 24}}{-4} \) \( x = \frac{-4 \pm \sqrt{40}}{-4} \) \( x = \frac{-4 \pm 2\sqrt{10}}{-4} \) We end up with two solutions for \( x \): \( x_1 = \frac{-4 + 2\sqrt{10}}{-4} \) and \( x_2 = \frac{-4 - 2\sqrt{10}}{-4} \) These are the x-intercepts of the parabola. **The Turning Point**: The turning point of a parabola in the form \( y = ax^2 + bx + c \) is given by the vertex of the parabola, at \( x = -\frac{b}{2a} \). For our parabola: \( x = -\frac{4}{2(-2)} \) \( x = -\frac{4}{-4} \) \( x = 1 \) Now, to find the y-coordinate of the turning point, we substitute \( x = 1 \) back into the equation: \( y = -2(1)^2 + 4(1) + 3 \) \( y = -2 + 4 + 3 \) \( y = 5 \) So the coordinates of the turning point (also known as the vertex of the parabola) are (1, 5). **Explaining why \( y = -2x^2 + 4x - 3 \) has two distinct real roots**: The graph of \( y = -2x^2 + 4x + 3 \) is a parabola that opens downwards due to the negative leading coefficient (-2). Since we know the parabola crosses the y-axis at (0, 3) and has a turning point at (1, 5) which is above the x-axis, and the parabola is symmetrical about the vertical line through the turning point, the graph must intersect the x-axis at two points. For the equation \( y = -2x^2 + 4x - 3 \), only the constant term is different. This would shift the parabola down by 6 units (from \( +3 \) in the original equation to \( -3 \) in the new equation). Since the original parabola already intersects the x-axis at two points, shifting it vertically downwards will not change the fact that it has two x-intercepts, but it will shift the position of these intercepts. Therefore, \( y = -2x^2 + 4x - 3 \) will also intersect the x-axis at two points, meaning it has two distinct real roots. However, without the actual graph provided, we're drawing these conclusions based on the characteristics of the quadratic function and the given coefficients.
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