Example Question - graph
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Understanding Functions and Graphs
<p>Để giải bài toán này, ta cần xác định tính chất của hàm số và đồ thị của nó.</p> <p>Câu 1: Xét hàm số \( y = ax^2 + bx + c \) với \( a \neq 0 \).</p> <p>A. Đoạn đồ thị \( P \) của hàm số này là parabol, có đỉnh tại \( \left( \frac{-b}{2a}, \frac{\Delta}{4a} \right) \).</p> <p>B. Đồ thị của hàm số này là \(\{ \frac{b}{a} - \frac{\Delta}{4a} \}\) với \(\Delta = b^2 - 4ac\).</p> <p>C. Giá trị của hàm số tại các điểm này được xác định bởi công thức trên.</p> <p>Câu 2: Tìm tập xác định \( D \) của hàm số \( y = \frac{3x-1}{2x-3} \) cần xác định điều kiện để mẫu khác không bằng 0.</p> <p>D = { x | 2x - 3 \neq 0 } tức là \( x \neq \frac{3}{2} \).</p>
Evaluating a Given Function Value from a Graph
<p>To find \( f(4) \), we locate the point where \( x = 4 \) on the graph and find the corresponding \( y \) value.</p> <p>At \( x = 4 \), the graph passes through the point \((4, 3)\).</p> <p>Therefore, \( f(4) = 3 \).</p>
Graph Interpretation and Vector Components
<p>Considérons les coordonnées des points A et B sur le graphique :</p> <p>Point A : (2, 3)</p> <p>Point B : (7, 1)</p> <p>Calculons les composantes du vecteur \(\overrightarrow{AB}\) :</p> <p>\(\Delta x = x_B - x_A = 7 - 2 = 5\)</p> <p>\(\Delta y = y_B - y_A = 1 - 3 = -2\)</p> <p>Les composantes du vecteur \(\overrightarrow{AB}\) sont donc (5, -2).</p> <p>Utilisons ces composantes pour trouver les vecteurs \(\overrightarrow{u}\) et \(\overrightarrow{v}\) :</p> <p>\(\overrightarrow{u}\) + \(\overrightarrow{v}\) = \(\overrightarrow{AB}\)</p> <p>\(\overrightarrow{u}\) + \(\overrightarrow{v}\) = (5, -2)</p> <p>Selon l'énoncé, \(\overrightarrow{u}\) = (2, a) et \(\overrightarrow{v}\) = (b, -3).</p> <p>Donc (2 + b, a - 3) = (5, -2).</p> <p>Égalons les composantes correspondantes :</p> <p>2 + b = 5 => b = 3</p> <p>a - 3 = -2 => a = 1</p> <p>Par conséquent, les valeurs de a et b sont respectivement 1 et 3.</p>
Transformation of Function Graphs
<p>К сожалению, изображение не содержит достаточно информации для предоставления точного решения. Чтобы решить эту задачу, мне нужно увидеть графики функций \( f(x) \) и \( g(x) \), а также знать, как они связаны через простое преобразование.</p> <p>Для части a) общие виды преобразований включают: 1. Сдвиг по вертикали: \( f(x) + k \) 2. Сдвиг по горизонтали: \( f(x + h) \) 3. Растяжение по вертикали: \( a \cdot f(x) \) 4. Растяжение по горизонтали: \( f(b \cdot x) \) 5. Отражение по вертикали: \( -f(x) \) 6. Отражение по горизонтали: \( f(-x) \)</p> <p>Для части b), чтобы построить \( y=f(x-2)-1 \), нужно сделать следующее: 1. Сдвинуть график \( f(x) \) на 2 единицы вправо по оси x. 2. Сдвинуть результат вниз на 1 единицу по оси y.</p>
Finding the Vertex of a Quadratic Function
<p>Para encontrar el vértice de la función cuadrática \( f(x) = 2x^2 - 8x \), podemos usar la fórmula del vértice para una parábola de la forma \( ax^2 + bx + c \), donde el vértice está dado por \( h = -\frac{b}{2a} \).</p> <p>En este caso, \( a = 2 \) y \( b = -8 \). Sustituyendo estos valores en la fórmula:</p> <p>\( h = -\frac{-8}{2 \cdot 2} = \frac{8}{4} = 2 \)</p> <p>Para encontrar el valor de \( k \), el cual es el valor de \( f(x) \) en \( h \):</p> <p>\( k = f(2) = 2(2)^2 - 8(2) = 2 \cdot 4 - 16 = 8 - 16 = -8 \)</p> <p>Así, el vértice de la parábola es \( (h, k) = (2, -8) \).</p>
Understanding Ohm's Law and Its Graphical Representation
<p>Ohm's Law states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points, and inversely proportional to the resistance (R) of the conductor. The mathematical equation representing Ohm's Law is \( V = IR \).</p> <p>To graphically represent Ohm's Law, one would typically plot voltage (V) on the vertical axis and current (I) on the horizontal axis. For a resistor with a constant resistance, the graph will be a straight line passing through the origin (0,0). This line indicates that as the voltage increases, the current through the resistor increases proportionally, and each point on the line stays consistent with Ohm's Law \( V = IR \). The slope of this line (rise over run) is equal to the resistance (R).</p>
Graphing a Linear Equation on a Coordinate Plane
<p>Para resolver esta pregunta, necesitamos graficar la ecuación lineal \(y = 3x - 2\) en el plano de coordenadas. Primero, utilizando la tabla de valores, podemos ver algunos puntos que ya se han calculado y que se pueden trazar en la gráfica.</p> <p>Paso 1: Trazar los puntos (3, 7), (2, 4) y (-2, -8) en el plano de coordenadas. Cada punto corresponde a un valor de 'x' y el valor de 'y' resultante tras aplicar la ecuación \(y = 3x - 2\).</p> <p>Paso 2: Dibujar una línea recta que pase por estos puntos, ya que representan la solución a la ecuación lineal y cualquier punto en esta línea satisfará la ecuación \(y = 3x - 2\).</p> <p>El punto donde la línea cruza el eje 'y' es el intercepto en y, que para esta ecuación es -2, y esto indica que cuando \(x=0\), \(y=-2\).</p> <p>La pendiente de la línea es 3, indicando que por cada aumento en 1 en la dirección de 'x', 'y' aumentará en 3 unidades.</p>
Vectors Representation Exercise
<p>The image seems to display graphical representation of vector addition and subtraction. However, without the context of the problem such as specific questions or instructions related with the vectors, no unique solution can be provided.</p> <p>The first box possibly shows two vectors being added head-to-tail.</p> <p>The second box might be showing vector subtraction by adding the negative of a vector.</p> <p>The third box possibly represents two equal vectors pointing in opposite directions, suggesting that their sum is zero.</p> <p>In the fourth box, it could be an example of the Parallelogram rule for vector addition.</p>
Determining the Domain and Range from a Graph
\[ \text{Domain: } \{ x \mid x < -4 \text{ or } x \geq 4\} \] \[ \text{Range: } \{ y \mid y \geq 4 \} \]
Coordinates of Madrid on a Graph
<p>El eje horizontal (x) representa los valores negativos a la izquierda del origen y los valores positivos a la derecha del origen.</p> <p>El eje vertical (y) representa los valores positivos por encima del origen y los valores negativos debajo del origen.</p> <p>Para encontrar la ubicación exacta de Madrid, buscamos el punto que representa a Madrid en la gráfica y luego determinamos sus coordenadas (x, y).</p> <p>Observando la gráfica, el punto para Madrid se encuentra en x = -3 y y = -4.</p> <p>Por lo tanto, las coordenadas ordenadas (x, y) para Madrid son (-3, -4).</p>
Sketching a Parametric Curve
\[ <p>1. Set up a table of values for t, x(t), and y(t).</p> <p>2. Calculate the corresponding x and y for at least 5 values of t within the interval \[-2, 2\].</p> <p>3. Plot the calculated points (x, y) on a coordinate plane.</p> <p>4. Draw the curve that connects these points smoothly, showing the movement as t increases.</p> \]
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.
Derivative Calculation with a Function and Point on a Graph
The image includes a graph displaying a straight line representing a function y = f(x), along with a pair of points labeled A(2, 1). Additionally, there's a mathematical question that states: Assuming the function g(x + 1) = (x^2 + 1) - f(x), what is the value of g'(3)? First, let's determine the slope of the line representing y = f(x) since we have the point A(2, 1) on the line. Since the line passes through the point A(2, 1) and the origin (0, 0), we can use these two points to calculate the slope (m): m = (y2 - y1) / (x2 - x1) m = (1 - 0) / (2 - 0) m = 1 / 2 Thus, the slope of the line is 1/2, and knowing that a line equation with slope m passing through the origin (0,0) has the form y = mx, the equation for f(x) is: f(x) = (1/2)x Now, let's put f(x) into the equation for g(x + 1): g(x + 1) = (x^2 + 1) - (1/2)x To find g'(x), the derivative of g(x), we must first write g(x) in terms of x: g(x) = ((x - 1)^2 + 1) - (1/2)(x - 1) g(x) = (x^2 - 2x + 1 + 1) - (1/2)x + (1/2) g(x) = x^2 - (5/2)x + 2 The next step is to differentiate g(x) with respect to x to find g'(x): g'(x) = 2x - 5/2 Finally, to find g'(3), substitute 3 into the derivative: g'(3) = 2(3) - 5/2 g'(3) = 6 - 5/2 g'(3) = (12/2) - (5/2) g'(3) = (12 - 5)/2 g'(3) = 7/2 Therefore, g'(3) is 7/2.
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