Example Question - x-intercepts
Here are examples of questions we've helped users solve.
Solving X-Intercepts of a Cubic Function
The image displays a math question with two sub-questions. Part a) is already solved in the image, where the given cubic function (x^3 - 6x^2 + 11x - 6) is factorized into (x - 1)(x - 2)(x - 3). For part b), we need to determine the x-intercepts of the cubic function. The x-intercepts of a function occur where the function equals zero. Since we have the factorized form of the cubic function, the x-intercepts can be found by setting each factor equal to zero and solving for x. Setting the factors equal to zero gives us: 1. x - 1 = 0, solving for x gives x = 1 2. x - 2 = 0, solving for x gives x = 2 3. x - 3 = 0, solving for x gives x = 3 Therefore, the x-intercepts are x = 1, x = 2, and x = 3. To sketch the graph, you would plot these x-intercepts on a coordinate axis. The cubic function is positive to the right of x = 3, negative between x = 2 and x = 3, positive between x = 1 and x = 2, and negative to the left of x = 1. The overall shape of the graph is an S-curve, with the ends of the curve heading upwards to the right and downwards to the left, considering it is a positive cubic polynomial. The turning points of the function will be between the x-intercepts, but without additional information, we cannot accurately determine their locations—just that they will be somewhere between the x-intercepts we have found.
Identifying Graph of Quadratic Function
To solve the question given in the image, which is to identify the graph of the quadratic function g(x) = -x^2 - 7x - 12, we should look at the properties of the quadratic function and check which graph corresponds to these. The given quadratic function is in the form g(x) = ax^2 + bx + c, where a = -1, b = -7, and c = -12. Because a is negative, we know that the parabola opens downward. The graph should therefore be a downward-opening parabola. Now let's find the x-intercepts (the roots of the quadratic equation) by factoring or using the quadratic formula. The quadratic formula is given by x = [-b ± sqrt(b^2 - 4ac)] / 2a. Using the coefficients from g(x) = -x^2 - 7x - 12 gives us: x = [-(-7) ± sqrt((-7)^2 - 4 * (-1) * (-12))] / 2 * (-1) x = [7 ± sqrt(49 - 48)] / -2 x = [7 ± sqrt(1)] / -2 x = [7 ± 1] / -2 This yields two solutions for x: x = (7 + 1) / -2 = 8 / -2 = -4 x = (7 - 1) / -2 = 6 / -2 = -3 So the x-intercepts are x = -4 and x = -3. By checking the options: - Option A shows a parabola that does not cross the x-axis at -4 and -3. - Option B shows a parabola with the correct roots of -4 and -3, but it is opening upward, which is incorrect. - Option C shows a parabola that opens downwards, but with the incorrect roots. - Option D shows a parabola opening downward with the correct roots of -4 and -3. Thus, the correct graph that corresponds to the function g(x) = -x^2 - 7x - 12 is the one in Option D.
Quadratic Function Analysis
The image shows a question with two parts, (i) and (ii), concerning the graph of the function \(y = -3x^2 + 4x + 3\). (i) **Find the intercepts and the coordinates of the turning point on the graph of** \(y = -3x^2 + 4x + 3\). **To find the x-intercepts**, we set \(y\) equal to zero and solve for \(x\): \[0 = -3x^2 + 4x + 3\] This is a quadratic equation, and we can solve it using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -3\), \(b = 4\), and \(c = 3\). Let's solve for \(x\): \(x = \frac{-4 \pm \sqrt{4^2 - 4(-3)(3)}}{2(-3)}\) \(x = \frac{-4 \pm \sqrt{16 + 36}}{-6}\) \(x = \frac{-4 \pm \sqrt{52}}{-6}\) \(x = \frac{-4 \pm 2\sqrt{13}}{-6}\) Since we cannot simplify the square root any further, we have: \(x_1 = \frac{-4 + 2\sqrt{13}}{-6}\) and \(x_2 = \frac{-4 - 2\sqrt{13}}{-6}\) **To find the y-intercept**, we set \(x\) equal to zero: \(y = -3(0)^2 + 4(0) + 3 = 3\) So the y-intercept is at the point (0, 3). **To find the turning point** (also known as the vertex), we can use the formula for the x-coordinate of the vertex, \(x = -\frac{b}{2a}\), where \(a = -3\) and \(b = 4\): \(x = -\frac{4}{2(-3)}\) \(x = \frac{4}{6}\) \(x = \frac{2}{3}\) Now substitute \(x = \frac{2}{3}\) into the equation to find the y-coordinate: \(y = -3\left(\frac{2}{3}\right)^2 + 4\left(\frac{2}{3}\right) + 3\) \(y = -3\left(\frac{4}{9}\right) + \frac{8}{3} + 3\) \(y = -\frac{4}{3} + \frac{8}{3} + 3\) \(y = \frac{4}{3} + 3\) \(y = \frac{4}{3} + \frac{9}{3}\) \(y = \frac{13}{3}\) So the coordinates of the turning point are \( \left(\frac{2}{3}, \frac{13}{3}\right) \). (ii) **Use the graph to explain why** \(y = -3x^2 + 4x + 3\) **has two distinct real roots.** The question is asking to use the graph as an explanation, but since we cannot see the graph, I will provide a mathematical explanation based on what we have calculated. The quadratic equation \(y = -3x^2 + 4x + 3\) has two distinct real roots because the square root part of the quadratic formula resulted in \(\sqrt{52}\), which is a real number (as all square roots of positive numbers are real). Having two distinct roots means that the discriminant \(b^2 - 4ac\) is positive (\(52\) in our case), and this leads to two different x-intercepts on the graph, which we calculated earlier. In the context of a graph, this means that the parabola crosses the x-axis at two points, corresponding to the two distinct real roots.
Analysis of a Parabola in Vertex Form
The equation provided in the image is a parabola in vertex form: \[ y = 3(x - 1)^2 + 4 \] To find the coordinates of the turning point (vertex) of the parabola, we can directly read them from the vertex form of the quadratic equation: \[ y = a(x - h)^2 + k \] where \((h, k)\) is the vertex of the parabola. In this equation, \(a = 3\), \(h = 1\), and \(k = 4\). Thus, the coordinates of the turning point (vertex) are: \[ (h, k) = (1, 4) \] To find the x-intercepts, set \(y = 0\) and solve for \(x\): \[ 0 = 3(x - 1)^2 + 4 \] \[ -4 = 3(x - 1)^2 \] \[ \frac{-4}{3} = (x - 1)^2 \] Since the square of a real number is always non-negative, the equation \((x - 1)^2 = -4/3\) has no real solutions. Thus, there are no x-intercepts for this parabola because it doesn’t cross the x-axis. To find the y-intercept, set \(x = 0\) and solve for \(y\): \[ y = 3(0 - 1)^2 + 4 \] \[ y = 3(1) + 4 \] \[ y = 3 + 4 \] \[ y = 7 \] So the y-intercept is at \(y = 7\) or at the point \((0, 7)\) on the graph. Lastly, the line of symmetry of a parabola is a vertical line that passes through the vertex. For the given equation, it is the line \(x = h\), which in this case is: \[ x = 1 \] This is the line of symmetry of the graph of the given quadratic function.
Equation of Rational Function with Given Conditions
To write an equation for a rational function that satisfies the given conditions, we can follow these steps: 1. Vertical asymptotes at \( x = 5 \) and \( x = -6 \) are given by factors in the denominator that become zero at these x-values. Thus, the denominator will be \( (x - 5)(x + 6) \). 2. X-intercepts at \( x = -5 \) and \( x = 1 \) indicate where the numerator of the function becomes zero. Thus, the numerator will be a product of factors that become zero at these x-values, which gives us \( (x + 5)(x - 1) \). 3. The y-intercept is the point where the function crosses the y-axis, which happens when \( x = 0 \). To get a y-intercept at 6, we need the constant term of the numerator (when \( x = 0 \)) to give us a y-value of 6. Since the denominator at \( x = 0 \) will be \( (0 - 5)(0 + 6) = -30 \), to obtain a y-intercept of 6, the numerator must evaluate to \( -30 * 6 \) when \( x = 0 \). Therefore, we can multiply the numerator by \( -6 \) to get the desired y-intercept. Putting it all together, the rational function is: \( y = \frac{-6(x + 5)(x - 1)}{(x - 5)(x + 6)} \) This function satisfies the given conditions: - The vertical asymptotes are at \( x = 5 \) and \( x = -6 \). - The x-intercepts are at \( x = -5 \) and \( x = 1 \). - The y-intercept is at \( y = 6 \) because when \( x = 0 \), \( y = \frac{-6(0 + 5)(0 - 1)}{(0 - 5)(0 + 6)} = \frac{30}{-30} = -1 \cdot -6 = 6 \).
Finding a Rational Function Given Vertical, X-intercepts, and Y-intercept
To write an equation for a rational function that meets the given requirements, you need to consider the vertical asymptotes, x-intercepts, and y-intercept provided. Vertical asymptotes occur where the denominator of the rational function is zero. Thus, for vertical asymptotes at \( x = -6 \) and \( x = 3 \), the factors in the denominator will be \( (x + 6) \) and \( (x - 3) \), since setting either of these equal to zero gives the respective values of \( x \) for the asymptotes. X-intercepts occur where the numerator is zero. Given the x-intercepts at \( x = -3 \) and \( x = -1 \), the factors in the numerator will be \( (x + 3) \) and \( (x + 1) \), since setting these factors equal to zero will yield the x-intercepts. Combining these, we get a function of the form: \[ f(x) = \frac{a(x + 3)(x + 1)}{(x + 6)(x - 3)} \] where \( a \) is a constant we need to determine using the y-intercept. The y-intercept is found by setting \( x = 0 \) in the function. It is given that the y-intercept is 10; thus, we solve for \( a \) by substituting \( x = 0 \) into the equation to get \( f(0) = 10 \): \[ 10 = \frac{a(0 + 3)(0 + 1)}{(0 + 6)(0 - 3)} \] \[ 10 = \frac{3a}{-18} \] Now solve for \( a \): \[ a = \frac{10 \times (-18)}{3} = -60 \] Insert \( a \) back into the original equation to get the complete function: \[ f(x) = \frac{-60(x + 3)(x + 1)}{(x + 6)(x - 3)} \]
Analyzing Quadratic Function and Finding Coordinates
To find the coordinates of A, B, and C, we need to analyze the given parabola and the quadratic function represented by the equation \( y = -(x - 7)(x + 1) \). First, let's expand this equation to find the standard form of the quadratic function: \[ y = -(x^2 + 1x - 7x - 7) \] \[ y = -(x^2 - 6x - 7) \] \[ y = -x^2 + 6x + 7 \] Now we will find the x-intercepts (A and C) by setting y to 0 and solving for x: \[ 0 = -x^2 + 6x + 7 \] Now we can factor the quadratic or use the quadratic formula to solve it. Since \( -x^2 + 6x + 7 \) can be factored easily, we will factor it: \[ 0 = -(x - 7)(x + 1) \] Setting each factor equal to zero gives us the x-intercepts: \[ x - 7 = 0 \implies x = 7 \] \[ x + 1 = 0 \implies x = -1 \] This gives us the coordinates of A and C: - A = (-1, 0) since the x-intercept is where y is zero. - C = (7, 0) for the same reason. Now, we need to find the vertex of the parabola (point B). The vertex form of a parabola is \( y = a(x - h)^2 + k \), where (h, k) is the vertex of the parabola. Since the given parabola is facing downwards (due to the negative sign in front of \( x^2 \)), the vertex is a maximum point. The x-coordinate of the vertex (h) of a parabola in standard form \( y = ax^2 + bx + c \) can also be found using the formula \( h = -\frac{b}{2a} \): Here, a = -1 and b = 6, so: \[ h = -\frac{b}{2a} \] \[ h = -\frac{6}{2(-1)} \] \[ h = -\frac{6}{-2} \] \[ h = 3 \] We can substitute x = 3 back into the expanded equation to find the y-coordinate of the vertex: \[ y = -x^2 + 6x + 7 \] \[ y = -3^2 + 6(3) + 7 \] \[ y = -9 + 18 + 7 \] \[ y = -9 + 25 \] \[ y = 16 \] So the coordinates of B, the vertex, are (3, 16). In conclusion, the coordinates of the points are: - A = (-1, 0) - B = (3, 16) - C = (7, 0)
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