Example Question - quadratic functions
Here are examples of questions we've helped users solve.
Analysis of Quadratic Functions
<p>Given the quadratic function \( y = ax^2 + bx + c \), we want to determine if the quadratic \( (P) \) can be formed from the parameters \( a \neq 0 \). This involves calculating the expression \( \frac{b \pm \sqrt{\Delta}}{2a} \), where \( \Delta = b^2 - 4ac \).</p> <p>To derive the vertex form, we need to ensure that \( \Delta \) is non-negative, leading to real roots that allow for the graphical representation of the quadratic function.</p>
Understanding Quadratic Functions
The equation given in the image is \( y = x^2 + 9 \). This equation represents a quadratic function, which is a type of nonlinear function. The nonlinearity is because the variable \( x \) is raised to the second power. Therefore, the correct answer to the question "Which statement explains the type of function that is represented by the equation \( y = x^2 + 9 \)?" is: D) The function is nonlinear because the variable \( x \) is raised to the second power.
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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