Example Question - maximum point
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Analysis of a Quadratic Function Graph
<p>To determine the coordinates of point \(Q\):</p> <p>Point \(Q\) lies on the x-axis, which means the y-coordinate is 0.</p> <p>Set the function \(f(x)=x^2+6x-5\) equal to 0 and solve for \(x\):</p> <p>\(x^2+6x-5=0\)</p> <p>\(x=\frac{-6\pm\sqrt{6^2-4(1)(-5)}}{2(1)}\)</p> <p>\(x=\frac{-6\pm\sqrt{36+20}}{2}\)</p> <p>\(x=\frac{-6\pm\sqrt{56}}{2}\)</p> <p>\(x=\frac{-6\pm 2\sqrt{14}}{2}\)</p> <p>\(x=-3\pm\sqrt{14}\)</p> <p>Since \(Q\) is to the right of the y-axis, we take the positive value:</p> <p>\(Q=(x, y)\)</p> <p>\(Q=(-3+\sqrt{14}, 0)\)</p> <p>To determine the maximum point \(P\):</p> <p>The vertex of a parabola \(y=ax^2+bx+c\) is given by the formula \(x=-\frac{b}{2a}\).</p> <p>In this function \(a=1\) and \(b=6\), so:</p> <p>\(x_P=-\frac{6}{2(1)}\)</p> <p>\(x_P=-3\)</p> <p>Substitute \(x_P\) into the function to find \(y_P\):</p> <p>\(y_P=(x_P)^2+6x_P-5\)</p> <p>\(y_P=(-3)^2+6(-3)-5\)</p> <p>\(y_P=9-18-5\)</p> <p>\(y_P=-14\)</p> <p>So the maximum point \(P\) is:</p> <p>\(P=(x_P, y_P)\)</p> <p>\(P=(-3, -14)\)</p>
Analyzing the Composition of a Given Function Graph
Given the function \( f(x) = 2x^3 - 9x^2 + 12x - 3 \), we are asked to determine its concavity, maximum point, and inflection point. To find the concavity of the function, we need to compute the second derivative and analyze its sign: \( f''(x) = \frac{d^2}{dx^2}(2x^3 - 9x^2 + 12x - 3) \) \( f''(x) = \frac{d}{dx}(6x^2 - 18x + 12) \) \( f''(x) = 12x - 18 \) To determine the concavity, we look at the sign of \( f''(x) \). If \( f''(x) > 0 \), the function is concave up. If \( f''(x) < 0 \), the function is concave down. So, the function is concave up when \( 12x - 18 > 0 \) i.e., \( x > \frac{3}{2} \), and concave down when \( x < \frac{3}{2} \). To find the maximum point, we need to set the first derivative equal to zero and solve for x: \( f'(x) = \frac{d}{dx}(2x^3 - 9x^2 + 12x - 3) \) \( f'(x) = 6x^2 - 18x + 12 \) Setting \( f'(x) = 0 \), we get: \( 6x^2 - 18x + 12 = 0 \) Dividing by 6: \( x^2 - 3x + 2 = 0 \) Factoring: \( (x - 2)(x - 1) = 0 \) Thus, \( x = 1 \) and \( x = 2 \) are critical points. We must check these points in the second derivative to see if they correspond to maximum points: \( f''(1) = 12(1) - 18 = -6 < 0 \), so \( x = 1 \) is a local maximum point. \( f''(2) = 12(2) - 18 = 6 > 0 \), so \( x = 2 \) is not a maximum point, it is a point of inflection since the concavity changes from down to up.
Analysis of Quadratic Function Graph Features
<p>To locate minimum Q, the coordinates of vertex \( Q \) of the parabola \( f(x) = ax^2 + bx + c \) can be found by using the vertex formula \( x = -\frac{b}{2a} \). The given equation can be rewritten as \( f(x) = -x^2 + 5 \), therefore we have \( a = -1 \) and \( b = 0 \).</p> <p>\[ x_Q = -\frac{b}{2a} = -\frac{0}{2(-1)} = 0 \]</p> <p>To find the y-coordinate of Q, substitute \( x_Q \) into \( f(x) \),</p> <p>\[ y_Q = f(x_Q) = -x_Q^2 + 5 = -(0)^2 + 5 = 5 \]</p> <p>Thus the coordinates of Q are \( (0,5) \).</p> <p>Given that the parabola opens downward (\( a < 0 \)), the vertex is the maximum point P. Therefore, the maximum point P is also at \( (0,5) \).</p>
Determine Coordinate Values and Maximum Point
<p>The image does not provide the necessary information or functions to calculate the specific coordinates or the maximum point as requested in the questions (a) and (b). Therefore, a general solution or approach cannot be provided without additional context or data from the original math problem.</p>
Solving for Coordinate Points and Identifying Extrema
Unfortunately, the image provided does not include any equations or enough context to solve for the coordinate point Q or to identify the maximum and minimum points of P. More information is needed to provide a solution. Please provide the necessary equations or additional context related to point Q and function P.
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