The quadratic function which takes the value 41 at \(x = -2\) and the value 20 at \(x = 5\) and is minimized at \(x = 2\) is …
\(f(x) = ax^2 + bx + c\)
\(f(-2) = 41\)
\(f(5) = 20\)
The function is minimized at \(x_p = 2\)
Use the vertex formula
\(f(x) = a(x\:-\:x_p)^2 + y_p\)
\(f(x) = a(x\:-\:2)^2 + y_p\)
From \(f(-2) = 41\)
\(41 = a(-2\:-\:2)^2 + y_p\)
\(41 = a(-4)^2 + y_p\)
\(41 = 16a + y_p\:\dotso \color{red} (1)\)
From \(f(5) = 20\)
\(20 = a(5\:-\:2)^2 + y_p\)
\(20 = a(3)^2 + y_p\)
\(20 = 9a + y_p \:\dotso \color{red} (2)\)
Subtracting the second equation from the first:
\((16a + y_p)\:-\:(9a + y_p) = 41\:-\:20\)
\(7a = 21\)
\(a = \dfrac{21}{7}\)
\(a = 3\)
Substituting \(a = 3\) into \(20 = 9a + y_p\)
\(20 = 9(3) + y_p\)
\(20 = 27 + y_p\)
\(y_p = -7\)
Substituting \(a = 3\) and \(y_p = -7\) into \(f(x) = a(x\:-\:2)^2 + y_p\)
\(f(x) = 3(x\:-\:2)^2 \:-\:7\)
\(f(x) = 3(x^2 \:-\:4x + 4) \:-\:7\)
\(f(x) = 3x^2 \:-\:12x + 12\:-\:7\)
\(f(x) = 3x^2 \:-\:12x + 5\)
Since the function is minimized at \(x_p = 2\), the minimum value of the function is \(y_p = -7\)
Conclusion:
The quadratic function which takes the value 41 at \(x = -2\) and the value 20 at \(x = 5\) and is minimized at \(x = 2\) is \(\color{red} f(x) = 3x^2 \:-\:12x + 5\)
The minimum value of this function is \(\color{red} -7\).
The equation of the parabola which moved the parabola A symmetrically with the origin is…
The given equaiton of the parabola is: \(y = x^2\:-\:4x \:-\:5\)
The vertex of a parabola \(y = ax^2 + bx + c\) is given by \(x_{\text{vertex}} = \dfrac{-b}{2a}\)
Substituting \(a = 1 \text{ and } b = -4\):
\(x_{\text{vertex}} = \dfrac{-(-4)}{2(1)} = \dfrac{4}{2} = 2\)
To find \(y_{\text{vertex}}\), substitute \(x = 2\) into the equation:
\(y = (2)^2 \:-\:4(2) \:-\:5\)
\(y = 4\:-\:8\:-\:5\)
\(y = -9\)
So, the vertex is \((2, -9)\)
To find the x-intercept, set y = 0
\(x^2 \:-\:4x \:-\:5 = 0\)
\((x\:-\:5)(x + 1) = 0\)
\(x\:-\:5 = 0 \rightarrow x_1 = 5\)
\(x + 1 = 0 \rightarrow x_2 = -1\)
So, the x-coordinates of the intersection point are: \(-1, 5\)
A reflection through the origin means replacing \(x\) with \(-x\) and \(y\) with \(-y\)
Starting from \(y = x^2\:-\:4x \:-\:5\)
Replace \(x\) with \(-x\) and \(y\) with \(-y\)
\(-y = (-x)^2 \:-\:4(-x) \:-\:5\)
\(-y = x^2 + 4x \:-\:5\)
\(y = -x^2 \:-\:4x + 5\)
Finding \(k\) when the line B: \(y = 2x + k\) touches the parabola A
Substituting \(y = 2x + k\) into \(y = x^2 \:-\:4x\:-\:5\)
\(2x + k = x^2 \:-\:4x \:-\:5\)
\(0 = x^2 \:-\:4x\:-\:2x \:-\:5\:-\:k\)
\(0 = x^2 \:-\:6x \:-\:(5 + k)\)
\(a = 1, b = -6, \text{ and } c = -(5 + k)\)
For the line to be tangent to the parabola, the discriminant must be zero:
\(b^2 \:-\:4ac = 0\)
\((-6)^2\:-\:4(1)(-5\:-\:k) = 0\)
\(36 + 20 + 4k = 0\)
\(56 = -4k\)
\(k = \dfrac{56}{-4}\)
\(k = -14\)
Conclusion:
The coordinate of the vertex of the parabola A is \(\color{red} (2, -9)\)
and the x-coordinate of the intersection of the parabola A and the x-axis is \(\color{red} -1, 5\)
The equation of the parabola which moved the parabola A symmetrically with the origin is \(\color{red} y = -x^2 \:-\:4x + 5\)
When a straight line B: \(y = 2x + k\) touches the parabola A, k = \(\color{red} -14\)
