Full Grade 9 test

Factor the numerator and denominator. Cancel only shared nonzero factors, never individual terms in a sum. Keep the original denominator restrictions even when cancellation changes the denominator.

An equation root is a value of x that makes the equation true when substituted for x.

For this question, you do not need to solve the whole equation. Substitute the given number for every x and calculate the left-hand side:

x² + bx + c.

1. Square the given number.
2. Multiply the given number by the coefficient b.
3. Add c and combine all the resulting numbers.

If the result is 0, the given number is a root of the equation. If the result is not 0, it is not a root.

When the roots of a quadratic equation must be found, first calculate the discriminant D = b² − 4ac, then use x = (−b ± √D) / 2a.

tan α = opposite leg / adjacent leg. Tangent relates both legs directly without using the hypotenuse. Use inverse tangent to find the angle.

Use the perfect-square identity:

(x + a)² = x² + 2ax + a².

Compare it with the given trinomial x² + kx + c:

• the constant term c must equal a²;
• the middle term kx must match 2ax.

Therefore, to find k, multiply a by 2. Substitute the result for k to check that the two expressions are identical.

At x = 0, all terms containing x vanish, so the y-intercept value is the constant term. In a linear cost formula it is the fixed fee; total cost = fixed fee + unit price × quantity. For a quadratic, the intercept is (0, c).

Factor out x: ax² + bx = x(ax + b) = 0. A product is zero when at least one factor is zero. Thus x = 0 or ax + b = 0, giving x = −b/a.

For c > 0, x² = c has two roots: x = −√c and x = √c. Include the negative root because a negative number also has a positive square.

From sin²α + cos²α = 1, obtain cos α = √(1 − sin²α), taking the positive root because α is acute. Square the sine first.

The vertex form of a parabola is y = a(x − m)² + n.

Its vertex coordinates can be read directly:

V(m, n).

Pay close attention to the sign inside the brackets. For example, if the expression is x − (−3), the x-coordinate of the vertex is −3, not 3.

First add AD and DB to find the whole side AB. Similar triangles ADE and ABC give DE / BC = AD / AB. DB is not the whole triangle side.

Find m and n whose sum is the x coefficient and whose product is the constant. Then x² + (m + n)x + mn = (x + m)(x + n). Check by expanding.

Each scatter-plot point represents a pair of numerical values. An upward trend indicates positive correlation, a downward trend negative correlation; no clear trend means no clear correlation.

Triangles AOB and COD have equal opposite angles at O; the parallel bases give another equal angle pair. Match corresponding sides and form a proportion using the bases and the diagonal segments.

The notation f(x) represents the function value obtained for a particular value of x.

1. Substitute the given number for every x.
2. Perform the multiplication first.
3. Then add the constant term.

If x or the constant term is negative, use brackets during substitution so that its sign is preserved.

The domain is the set of allowed input values x. A function assigns exactly one y to each x in its domain. Different inputs may share an output, but one input cannot have two different outputs.

Find the full tree shadow length, rather than using only the difference. Heights and shadows form similar right triangles, so the height ratio equals the shadow length ratio.

The graph of y = ax² is a parabola. The value |a| shows how strongly the parabola is horizontally compressed or stretched.

• The larger |a| is, the narrower the parabola becomes.
• The closer |a| is to zero, the wider the parabola becomes.

The sign of a determines its direction: when a > 0, the parabola opens upwards; when a < 0, it opens downwards.

Each next term in a geometric sequence is the preceding term multiplied by the constant ratio q. Find q by dividing consecutive terms. The nth term is aₙ = a₁ × qⁿ⁻¹.

Percentage grade = vertical rise ÷ horizontal distance × 100. A 12% grade means a 12 m rise over 100 m horizontally. It is not a 12° angle.

A graph crosses the x-axis where its y-coordinate equals zero.

For y = ax + b, substitute y = 0 and solve:

ax + b = 0, so x = −b/a.

The intersection point is written as (x, 0). Check the answer by substituting the calculated x into the original function.

Take the square root of the area ratio to find the length ratio. Use the vertex order to match BC to its corresponding side, then apply that ratio.

The angle of elevation is measured from the horizontal to the upward line of sight. Do not measure it from the vertical wall.

Rearrange aₙ = a₁ + (n − 1)d to n = (aₙ − a₁)/d + 1. Subtract the first term from the given value, divide by the common difference and add 1.

The denominator product must be nonzero, so neither factor may be zero. From (x − a)(x − b), exclude x = a and x = b.

The perimeter ratio equals the length ratio. Areas scale by the square of that ratio, rather than by the length factor alone.

Replace division by a fraction with multiplication by its reciprocal:

A/B ÷ C/D = A/B · D/C.

Then factor the difference of squares:

x² − a² = (x − a)(x + a).

Cancel matching factors. Remember that the divisor cannot equal zero, so in addition to denominator restrictions you must exclude the value that makes the numerator of the second fraction zero.