The entire graphs of the equation y = x² + kx - x + 9 is strictly above the x-axis if and only if (a

Q: Wait, why do we need $D < 0$ instead of $D \le 0$?

That is a fantastic question! If $D = 0$, the parabola touches the x-axis at exactly one point (the vertex). The problem states the graph is 'strictly above' the x-axis, which means $y > 0$ for all $x$. If $D = 0$, there would be a point where $y = 0$, which is not 'strictly above'. Thus, we must have $D < 0$ to ensure the parabola never touches or crosses the x-axis.

Q: Why can't we just use energy conservation or some other physics principle?

I love the enthusiasm! However, this is a pure mathematical condition problem. In physics, we often use energy conservation to find motion, but here, we are analyzing the *existence* of a geometric property. We are essentially asking for the range of a parameter $k$ that keeps the entire function positive. The discriminant is the most powerful tool in our algebraic toolkit for this specific task.

Q: What if the coefficient of $x^2$ was negative? Would the condition change?

Absolutely! If the coefficient of $x^2$ (which is $a$) were negative, the parabola would open downwards. A downward-opening parabola will *always* eventually cross the x-axis, no matter where its vertex is. Therefore, it could never be 'strictly above' the x-axis for all $x$. So, the condition $a > 0$ is just as vital as $D < 0$.

Question

The entire graphs of the equation $y = x^2 + kx - x + 9$ is strictly above the x-axis if and only if

Options:

(A) $k < 7$

(B) $-5 < k < 7$

(C) $k > -5$

(D) None of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing on a vast, flat plain—the $x$-axis. Above you, a parabola is suspended in the air. It is not touching the ground; it is not even grazing it. It is floating, completely above the $x$-axis, for every single value of $x$ you can imagine. This is the geometric reality of the problem we are solving today: $y = x^2 + kx - x + 9$. We want to find the values of $k$ that keep this parabola 'floating' in the positive region of the Cartesian plane. ### The Two Pillars of Stability To ensure a quadratic equation $y = ax^2 + bx + c$ stays strictly above the $x$-axis, we need to satisfy two fundamental conditions. Think of these as the two pillars holding up our floating parabola. First, the parabola must open upwards. If it opened downwards, it would eventually dive towards negative infinity, crossing the $x$-axis and violating our condition. This means our leading coefficient, $a$, must be strictly positive: $a > 0$. In our specific equation, $y = x^2 + (k-1)x + 9$, the coefficient $a$ is $1$. Since $1 > 0$, our first pillar is rock solid. The parabola is indeed a 'cup' shape. Second, the parabola must not touch the $x$-axis. If it touched the axis, the $y$-value would be zero at that point, and we need $y > 0$ everywhere. This brings us to the discriminant, $D = b^2 - 4ac$. The discriminant is the DNA of a quadratic equation; it tells us everything about its roots. If $D > 0$, the parabola crosses the axis twice. If $D = 0$, it kisses the axis once. But if $D < 0$, the parabola has no real roots—it never touches the $x$-axis. This is exactly what we need! ### The Algebraic Journey Let's refine our equation. We are given $y = x^2 + kx - x + 9$. Let's group the terms involving $x$ to reveal the true structure: $y = x^2 + (k-1)x + 9$. Now, we can clearly identify our coefficients: $a = 1$, $b = (k-1)$, and $c = 9$. Now, we apply our second condition: $D < 0$. Substituting our values into the discriminant formula, we get: $$(k-1)^2 - 4(1)(9) < 0$$ This simplifies to $(k-1)^2 - 36 < 0$. Instead of expanding the square, let's use the elegance of the difference of squares identity: $A^2 - B^2 = (A-B)(A+B)$. We know that $36$ is $6^2$. So, our inequality becomes: $$(k-1)^2 - 6^2 < 0$$ Applying the identity, we get: $$((k-1) - 6)((k-1) + 6) < 0$$ Simplifying inside the brackets, we arrive at: $$(k - 7)(k + 5) < 0$$ ### The Final Resolution We are left with a simple quadratic inequality. We have two critical points: $k = 7$ and $k = -5$. For the product of two factors to be negative, the value of $k$ must lie between the two roots. If $k$ were greater than $7$, both factors would be positive. If $k$ were less than $-5$, both factors would be negative, making the product positive. Therefore, the only way for the product to be negative is if $k$ is trapped between $-5$ and $7$. Thus, the range of $k$ is **$-5 < k < 7$**. We have successfully constrained the parameter $k$ to keep our parabola floating gracefully above the $x$-axis.

Animated Solution for Mathematics - Quadratic Equations: The entire graphs of the equation $y = x^{2} + k x - x + 9$ is strictly above the x-axis if and only if

Select Answer:

Visualized Solution

Visualizing the Problem

Geometric Interpretation

Mathematical Conditions

Standardizing the Equation

Identifying Coefficients

Checking Condition 1: $a > 0$

Setting up Condition 2: $D < 0$

Simplifying the Inequality

Factorizing the Expression

Simplifying the Factors

Solving the Inequality

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

Analyzing the Setup

The Two Pillars of Stability

The Algebraic Journey

The Final Resolution

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The entire graphs of the equation y=x2+kx−x+9 is strictly above the x-axis if and only if