Chapter 5, Problem 6CR

(a) Graph the quadratic function $f(x)=-x^2+2x-3$ by determining whether its graph is

concave up or concave down and by finding its vertex, axis of symmetry, $y$-intercept,

and $x$–intercept(s), if any.

(b) Solve $f(x)\le 0$.

To determine

To graph: The quadratic function $f\left(x\right)=-x^2+2x-3$ by determining whether its graph is concave up or down, and by finding vertex, axis of symmetry, $y-$ intercept, and $x-$ intercepts, if any.

Explanation of Solution

Given information:

The quadratic function $f\left(x\right)=-x^2+2x-3$.

Explanation:

Comparing equation $f\left(x\right)=-x^2+2x-3$ with $f\left(x\right)=a{x}^2+bx+c$ gives $a=-1,b=2,c=-3$.

The graph of quadratic equation $f\left(x\right)=a{x}^2+bx+c$ is concave up when leading coefficient $a>0$, and it is concave down when leading coefficient $a<0$.

In an equation $f\left(x\right)=-x^2+2x-3$, $a=-1<0$. So, graph of $f\left(x\right)=-x^2+2x-3$ would be concave downward.

Vertex of $f\left(x\right)=a{x}^2+bx+c$ is $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$.

In equation $f\left(x\right)=-x^2+2x-3$, $-\frac{b}{2a}=-\frac{2}{2\left(-1\right)}$

$\Rightarrow -\frac{b}{2a}=-\frac{2}{-2}$

$\Rightarrow -\frac{b}{2a}=1$

To find $f\left(-\frac{b}{2a}\right)$, that is, to find $f\left(1\right)$, substitute $x=1$ in $f\left(x\right)=-x^2+2x-3$.

$\Rightarrow f\left(1\right)=-{\left(1\right)}^2+2\left(1\right)-3$

$\Rightarrow f\left(1\right)=-1+2-3$

$\Rightarrow f\left(1\right)=-2$

Therefore, vertex of $f\left(x\right)=-x^2+2x-3$ is $\left(1,-2\right)$.

For an equation $f\left(x\right)=a{x}^2+bx+c$, axis of symmetry is line $x=-\frac{b}{2a}$.

Therefore, axis of symmetry for $f\left(x\right)=-x^2+2x-3$ would be line $x=1$.

To find $y-$ intercept of $f\left(x\right)=-x^2+2x-3$, plug $x=0$,

$\Rightarrow y=-{\left(0\right)}^2+2\left(0\right)-3$

$\Rightarrow y=-3$

Therefore, $y-$ intercept of $f\left(x\right)=-x^2+2x-3$ is $\left(0,-3\right)$.

Since $f\left(x\right)=-x^2+2x-3$ is concave downward parabola, vertex point is maximum point. Maximum point of $f\left(x\right)=-x^2+2x-3$ is $\left(1,-2\right)$, which is below $x-$ axis. That means, the graph of function $f\left(x\right)=-x^2+2x-3$ does not intersect with $x-$ axis. So, there would be no $x-$ intercept for function $f\left(x\right)=-x^2+2x-3$.

Plot the points $\left(1,-2\right),\left(0,-3\right)$ and line $x=1$ on $xy-$ plane, and draw the parabola symmetric about line $x=2$ and passing through points $\left(1,-2\right),\left(0,-3\right)$.

Therefore, graph of this function $f\left(x\right)=-x^2+2x-3$ is:

Interpretation:

The graph represents the quadratic function $f\left(x\right)=-x^2+2x-3$, withvertex $\left(1,-2\right)$, $y-$ intercept $\left(0,-3\right)$ and axis of symmetry $x=1$.

To determine

The solution of $f\left(x\right)\le 0$, where $f\left(x\right)=-x^2+2x-3$.

Answer to Problem 6CR

Solution:

Thesolution for $f\left(x\right)\le 0$ is all real numbers, that is, $\underset{\_}{\left(-\infty ,\infty \right)}$.

Explanation of Solution

Given information:

The function $f\left(x\right)=-x^2+2x-3$.

Explanation:

Solving the inequality $f\left(x\right)\le 0$ means finding $x$’s so that graph of function $f\left(x\right)$ is below $x-$ axis.

From graph of $f\left(x\right)=-x^2+2x-3$ in part (a), function $f\left(x\right)$ is below $x-$ axis for every value of $x.$

Therefore, the solution of the inequality $f\left(x\right)\le 0$ is all real numbers.

In interval notation, solution of $f\left(x\right)\le 0$ is $\left(-\infty ,\infty \right)$.