Sorry. I made a typo. Here's the correct one.Use discriminant to find

1.	Sorry. I made a typo. Here's the correct one.Use discriminant to find the minimum value of and explain the method.Hint: 'intersects' means discriminant is above zero.
Answer» Solution First things first, $y=f(x)$ is a parabola $y=k$ is a horizontal line If TWO intersects, that will be the solution of the SYSTEM EQUATION of $\begin{cases} & y=k \\ & y=f(x) \end{cases}$ So let's solve the system equation. $f(x)=k$ $\longleftrightarrow 2x^2-7x+2=k$ $\longleftrightarrow 2x^2-7x-k+2=0$ $\boxed{D=33+8k\geq 0}$ Thus, $y=k$ intersects if and only if $k\geq -\dfrac{33}{8}$ . So, the LOWEST location of the horizontal line is $y=-\dfrac{33}{8}$ , so it is the minimum value. More INFORMATION $y=f(x)$ is minimum when $x=\dfrac{7}{4}$ , because it is the axis of symmetry. View attachment for the explanation.

Sorry. I made a typo. Here's the correct one.Use discriminant to find the minimum value of and explain the method.Hint: 'intersects' means discriminant is above zero.

Answer»

Solution

First things first,

$y=f(x)$ is a parabola
$y=k$ is a horizontal line

If TWO intersects, that will be the solution of the SYSTEM EQUATION of

$\begin{cases} & y=k \\ & y=f(x) \end{cases}$

So let's solve the system equation.

$f(x)=k$

$\longleftrightarrow 2x^2-7x+2=k$

$\longleftrightarrow 2x^2-7x-k+2=0$

$\boxed{D=33+8k\geq 0}$

Thus, $y=k$ intersects if and only if $k\geq -\dfrac{33}{8}$ . So, the LOWEST location of the horizontal line is $y=-\dfrac{33}{8}$ , so it is the minimum value.