Mathematics IV (Advanced Algebra,Trigonometry And Statistics)

The statement is true because a parabola has a line of symmetry that divides it into two symmetrical parts. This line of symmetry acts as a mirror, reflecting one side of the parabola onto the other side. This property of a parabola can be observed by folding a piece of paper along the line of symmetry, where the two halves will perfectly overlap each other.

The given expression is a quadratic trinomial in the form of x^2 - 64. This can be factored using the difference of squares formula, which states that a^2 - b^2 can be factored as (a + b)(a - b). In this case, a = x and b = 8. Therefore, the factored form of the expression is (x + 8)(x - 8).

The turning point of a parabola is the vertex. The vertex is the point on the parabola where it changes direction from upward to downward or vice versa. It is the highest or lowest point on the graph of the parabola, depending on whether the parabola opens upward or downward. The vertex is also the point where the parabola intersects the line of symmetry, which is a vertical line that divides the parabola into two symmetrical halves. Therefore, the vertex is the correct answer.

The given expression (x+2)(x+2) or (x+2)^2 represents the square of the binomial (x+2). To find the product, we use the FOIL method, which stands for First, Outer, Inner, Last. When we apply the FOIL method, we multiply the first terms of both binomials (x and x), the outer terms (x and 2), the inner terms (2 and x), and the last terms (2 and 2). Simplifying these products, we get x^2+4x+4, which is the correct answer.

The trinomial x^2-8x+16 is a perfect square trinomial because it can be factored into (x-4)(x-4), which is equivalent to (x-4)^2. This shows that it is the square of a binomial, making it a perfect square trinomial.

To make the given trinomial a perfect square, we need to find the square of half the coefficient of the 'a' term. In this case, half of -14a is -7a. The square of -7a is 49a^2. Therefore, we need to add 49a^2 to the trinomial to make it a perfect square. Since there is no variable term in the trinomial, we can simply say that we need to add 49 to make it a perfect square trinomial.

The function f(x) = x-2 is not quadratic because it does not have a squared term. In a quadratic function, the highest power of the variable should be 2. In this case, the highest power of x is 1, so it is a linear function.

The statement is true because the vertex of a quadratic function in the form f(x) = ax^2 is always at the origin (0,0). This is because the vertex is the point on the graph where the function reaches its minimum or maximum value, and for a quadratic function, the minimum or maximum value is always at the vertex. Since the coefficient of x^2 is a, the function is symmetric about the y-axis, and the vertex is located at the origin.

When a parabola opens downward, the vertex is actually the maximum point, not the minimum point. The vertex is the point where the parabola reaches its highest or lowest point, depending on whether it opens upward or downward. Therefore, the given statement is incorrect.

The given trinomial y^2+8y+7 is not a perfect square trinomial because it cannot be factored into two identical binomials. In a perfect square trinomial, the first and last terms are perfect squares, and the middle term is twice the product of the square root of the first term and the square root of the last term. However, in y^2+8y+7, the first and last terms (y^2 and 7) are not perfect squares, and the middle term (8y) does not meet the criteria for a perfect square trinomial.

The solution set of the equation x^2+6x-7 is 1 and -7. This can be found by factoring the equation as (x+7)(x-1) = 0 and setting each factor equal to 0 to solve for x. Therefore, x = 1 and x = -7 are the values that satisfy the equation.

The vertex of a parabola in the form f(x) = a(x-h)^2 + k is located at the point (h, k). In this case, the equation is f(x) = 3(x-1)^2 +4, so the vertex is at (1, 4).

When a parabola opens upward, the vertex is the minimum point, not the maximum point. The vertex is the point where the parabola reaches its highest or lowest point depending on the direction it opens. Therefore, the correct answer is False.

The given statement is false. A polynomial of degree 2 is not called a square root function. A square root function is a type of function that involves finding the square root of a number. A polynomial of degree 2 is a quadratic function, which can be written in the form of ax^2 + bx + c, where a, b, and c are constants.

The axis of symmetry of a quadratic function is a vertical line that passes through the vertex of the parabola. In this case, the function is in the form f(x) = (x+1)^2 + 1, which represents a parabola that opens upwards. The vertex of the parabola is at (-1, 1), which means the axis of symmetry is a vertical line passing through x = -1. Therefore, the correct answer is x = -1.

The quadratic function f(x)=4x^2 has the narrowest graph because the coefficient of x^2 is the largest among all the given options. The coefficient of x^2 determines the steepness of the graph. A larger coefficient results in a steeper graph, while a smaller coefficient results in a wider graph. Therefore, f(x)=4x^2 has the narrowest graph compared to the other options.

To find the product of (6x-3) and (x+5), we can use the distributive property. We multiply each term in the first expression (6x-3) by each term in the second expression (x+5).

(6x * x) + (6x * 5) + (-3 * x) + (-3 * 5)
= 6x^2 + 30x - 3x - 15
= 6x^2 + 27x - 15

Therefore, the correct answer is 6x^2 + 27x - 15.

The quadratic function f(x)=4x^2 has the vertex at the origin because the vertex of a quadratic function in the form f(x)=ax^2+bx+c is given by (-b/2a, f(-b/2a)). In this case, a=4, b=0, and c=0. Therefore, the x-coordinate of the vertex is -b/2a = 0, and the y-coordinate is f(0) = 0. Hence, the vertex is at the origin (0,0).

The quadratic function with zeros 2 and -3 can be found using the fact that the zeros of a quadratic function are the values of x for which the function equals zero. In this case, the zeros are 2 and -3. To find the quadratic function, we can use the factored form of a quadratic equation, which is (x - r1)(x - r2) = 0, where r1 and r2 are the zeros. Plugging in the given zeros, we get (x - 2)(x + 3) = 0. Expanding this equation gives us x^2 + x - 6, which matches the first option, f(x) = x^2 + x - 6.

The equation of a quadratic function with a vertex at (0,3) can be written in the form f(x) = a(x-h)^2 + k, where (h,k) represents the vertex. In this case, h=0 and k=3. Plugging in these values, we get f(x) = a(x-0)^2 + 3, which simplifies to f(x) = ax^2 + 3. Since there are no other terms in the equation, the correct answer is f(x) = x^2 + 3.

The given quadratic equation is in the form of ax^2 + bx + c = 0. By applying the quadratic formula, x = (-b ± √(b^2 - 4ac)) / 2a, we can find the roots. In this case, a = 1, b = -5, and c = -6. Plugging these values into the quadratic formula, we get x = (5 ± √(25 + 24)) / 2. Simplifying further, we have x = (5 ± √49) / 2. This gives us two possible solutions: x = (5 + 7) / 2 = 6 and x = (5 - 7) / 2 = -1. Therefore, the roots of the equation are 6 and -1.

The equation y=(x-2)^2 represents a parabola that has its axis of symmetry shifted 2 units to the right of the y-axis. This can be determined by the presence of the term (x-2) in the equation, which indicates that the vertex of the parabola is located at x=2. Since the vertex is not on the y-axis, this parabola does not have its axis of symmetry on the y-axis.

The equation y=1/4x^2 represents a parabola with a coefficient of 1/4 in front of the x^2 term. The coefficient determines the width of the parabola, with a smaller coefficient resulting in a wider graph. Since 1/4 is the smallest coefficient among the given options, the equation y=1/4x^2 has the widest graph.

The given quadratic equation 3x^2+5x+1=0 has real roots because the discriminant (b^2-4ac) is positive. Since the discriminant is positive and not a perfect square, the roots are irrational. Additionally, the roots are unequal because the quadratic equation does not have a perfect square trinomial form. Therefore, the nature of the roots of the given equation is real, unequal, and irrational.

The quadratic function f(x)=-x^2+3 has a maximum point because the coefficient of x^2 is negative. When the coefficient of x^2 is negative, the graph of the quadratic opens downwards, creating a concave shape. In this case, the maximum point occurs at the vertex of the parabola, which is the highest point on the graph. Therefore, f(x)=-x^2+3 has a maximum point.