Find range of function $f(x)=x^2+6x-1$ for domain $x \in \mathbb{R}$.
The solution is:
Let $y=x^2+6x-1$ then $0=x^2+6x-1-y$ comparing with $0=ax^2+bx+c$ gives $a=1,b=6,c=-(1+y)$ For real $x$, $b^2-4ac \geq 0$ i.e $6^2 -(4)(1)[-(1+y)] \geq 0$ giving $y \geq -10$
Thus for real $x$ the function $f(x)=x^2+6x-1$ has range $\{ y \in \mathbb{R}: y \ge1 -10 \}$
I don't understand the technique of using discriminant to find range of function. I need assistance.
On
As mentioned in the comments, one way to find the range is to complete the square. \begin{align*} f(x) & = x^2 + 6x - 1\\ & = (x^2 + 6x) - 1\\ & = (x^2 + 6x + 9) - 9 - 1\\ & = (x + 3)^2 - 10 \end{align*} which is the equation of a parabola with vertex $(-3, -10)$ that opens upwards. Hence, the range of the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2 + 6x - 1$ is $[-10, \infty)$.
If $y$ is in the range of the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2 + 6x - 1$ then it must satisfy the quadratic equation $$x^2 + 6x - 1 = y$$ or, equivalently, $$x^2 + 6x - 1 - y = 0$$ We know that a quadratic equation has real roots if its discriminant is nonnegative. In this case, the discriminant is \begin{align*} \Delta & = b^2 - 4ac\\ & = 6^2 - 4(1)(-1 - y)\\ & = 36 + 4 + 4y\\ & = 40 + 4y \end{align*} Hence, $y$ is in the range of the function $f: \mathbb{R} \to \mathbb{R}$ if \begin{align*} \Delta & \geq 0\\ 40 + 4y & \geq 0\\ 4y & \geq -40\\ y & \geq -10 \end{align*}
The problem is to determine which the numbers $y$ such that the equation $x^2+6x-1=y$, which is equivalent to $x^2+6x-1-y=0$. But this quadratic equation has solutions if and only if $6^2+4(1+y)\ge0$, and $$ 6^2+4(1+y)\ge0\iff1+y\ge-9\iff y\ge-10. $$