Quadratic Functions Practice Quiz

The standard form of a quadratic equation is ax^2 + bx + c = 0, which clearly identifies the coefficients for the quadratic, linear, and constant terms. This form is most useful when applying methods such as the quadratic formula.

A quadratic equation can have at most two real roots. When the discriminant is positive, there are two distinct real roots, making 2 the maximum number.

Quadratic functions graph as parabolas that open either upward or downward depending on the leading coefficient. This shape distinguishes them from linear, circular, or hyperbolic graphs.

The axis of symmetry is found using the formula x = -b/(2a). For the quadratic y = 2x^2 + 4x + 1, substituting a = 2 and b = 4 gives x = -4/4, which simplifies to x = -1.

The vertex form of a quadratic function is given by y = a(x - h)^2 + k, where (h, k) is the vertex of the parabola. This form is particularly useful for graphing as it directly shows the vertex and the direction in which the parabola opens.

The vertex of a quadratic function is located at x = -b/(2a). For y = x^2 - 6x + 5, this gives x = 3. Substituting x = 3 into the equation, we find the y-coordinate to be -4, yielding the vertex (3, -4).

The discriminant, calculated as b^2 - 4ac, reveals the nature of the roots of a quadratic equation. A positive discriminant indicates two distinct real roots, while zero yields one repeated root and a negative results in complex roots.

The quadratic x^2 + 5x + 6 factors neatly into (x + 2)(x + 3) because 2 and 3 add up to 5 and multiply to 6. This classic example demonstrates the process of factoring by finding two numbers with the appropriate sum and product.

The equation x^2 - x - 12 = 0 factors into (x - 4)(x + 3) = 0. Setting each factor equal to zero yields the solutions x = 4 and x = -3.

By completing the square for x^2 + 6x, one adds and subtracts 9 to form (x + 3)^2, then combines the constant terms to get -4. This transforms the equation into vertex form: y = (x + 3)^2 - 4, revealing the vertex clearly.

The coefficient a controls the "width" or the vertical stretch of the parabola. A larger absolute value of a makes the graph steeper and narrower, while a smaller value results in a wider parabola.

Applying the quadratic formula to 2x^2 - 4x - 6 = 0, the discriminant is calculated as 64, leading to the solutions x = (4 ± 8)/4. This simplifies to x = 3 and x = -1.

The axis of symmetry for a quadratic is given by x = -b/(2a). For y = -3x^2 + 6x - 1, substituting a = -3 and b = 6 results in x = 1. This line divides the parabola into two mirror images.

Substituting the point (3, 1) into the vertex form yields 1 = a(1)^2 - 3, which simplifies to a = 4. This value determines the vertical stretch of the parabola.

The discriminant is computed as b^2 - 4ac, which is a critical component in determining the nature of the roots. Its value indicates whether the quadratic equation has two real roots, one repeated root, or two complex roots.

By substituting the given points into y = ax^2 + bx + c, a system of equations is formed. Solving this system yields a = 1, b = 2, and c = 1, so the quadratic function is y = x^2 + 2x + 1, which factors into (x + 1)^2.

If the graph of a quadratic function is tangent to the x-axis, it touches the axis at exactly one point. This happens when the discriminant (b^2 - 4ac) is zero, resulting in one repeated real root.

Starting with the factored form y = a(x + 2)(x - 5) and substituting the point (0, 10) gives 10 = a(2)(-5), which simplifies to a = -1. Thus, the quadratic function is y = - (x + 2)(x - 5).

Writing the equation in vertex form gives y = a(x - 3)^2 + 7. By substituting the point (5, 3) into this equation, we solve for a and find that a = -1. This confirms the parabola opens downward, and the correct vertex form is y = -(x - 3)^2 + 7.

The sign of the coefficient a is crucial in determining the direction the parabola opens. A positive value of a results in the parabola opening upward, while a negative value causes it to open downward.