To evaluate a function, simply substitute the value to the expression.

$f(x)=-3x^2+4x-1$

a) $f(1)=-3(1)^2+4(1)-1=0$

b) $f(-2)=-3(-2)^2+4(-2)-1=-21$

c) $fleft(dfrac{1}{3}right)=-3left(dfrac{1}{3}right)^2+4left(dfrac{1}{3}right)-1=0$

d) $f(0)=-3(0)^2+4(0)-1=-1$

e) $f(k)=-3k^2+4k-1$

f) $f(-k)=-3(-k)^2+4(-k)-1=-3k^2-4k-1$

a) 0
b) $-$21
c) 0
d) $-$1
e) $-3k^2+4k-1$
$f) -3k^2-4k-1$

$x^2+5x-3x-15$

$$
x^2+2x-15
$$

$$
2x^2+12x
$$

$-3(x+2)(x+2)+3$

$-3(x^2+2x+2x+4)+3$

$-3(x^2+4x+4)+3$

$-3x^2-12x-12+3$

$$
-3x^2-12x-9
$$

$(x-1)(x-1)$

$x^2-x-x+1$

$$
x^2-2x+1
$$

For a parabola $y=(x-h)^2+k$

$bold{vertex}$ $(h,k)$ is the point corresponding to the maximum or minimum value.

$bold{axis;of;symmetry}$, $x=h$ is the line that divides the parabola into two equal parts

$bold{domain}$, set of all possible values of $x$ which is $bold{R}$ for all quadratic functions.

$bold{range}$, set of all possible values of $y$

Refer to the graph provided in your textbook.

a) vertex$(-3,-4)$

axis of symmetry: $x=-3$

domain: ${ xin bold{R}}$

It opens downward, so $y$ values can’t be more than that of the vertex

range: ${ yin bold{R};|;yleq -4}$

b) vertex$(5,1)$

axis of symmetry: $x=5$

domain: ${ xin bold{R}}$

It opens upward, so $y$ values can’t be less than that of the vertex

range: ${ y in bold{R};|;ygeq 1}$

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|l|}
hline
& a) & b) \ hline
vertex & $(-3,-4)$ & $(5,1)$ \ hline
axis of symmetry & $x=-3$ & $x=5$ \ hline
domain & ${ xin bold{R}}$ & ${ xin bold{R}}$ \ hline
range & ${yin bold{R};|;yleq -4}$ & ${yin bold{R};|;ygeq 1}$ \ hline
end{tabular}
end{table}

a) In the form $y=a(x-h)^2+k,$ the given equation, $y=x^2+4
,$ is equivalent to

$$
begin{align*}
y&=(x-0)^2+4
.end{align*}
$$

Since the vertex of the quadratic equation in the form $y=a(x-h)^2+k$ is given by $(h,k),$ then the vertex is $left(
0,4
right).$

Since the axis of symmetry of a quadratic equation is given by $x=h,$ then the axis of symmetry is

$$
begin{align*}
x=0
.end{align*}
$$

Since in the given equation, $a>0$ ($a=
1
,$) then the parabola defined by the given equation opens
up.

b) In the form $y=a(x-h)^2+k,$ the given equation, $y=3(x-4)^2+1
,$ is equivalent to

$$
begin{align*}
y&=3(x-4)^2+1
.end{align*}
$$

Since the vertex of the quadratic equation in the form $y=a(x-h)^2+k$ is given by $(h,k),$ then the vertex is $left(
4,1
right).$

Since the axis of symmetry of a quadratic equation is given by $x=h,$ then the axis of symmetry is

$$
begin{align*}
x=4
.end{align*}
$$

Since in the given equation, $a>0$ ($a=
3
,$) then the parabola defined by the given equation opens
up.

c) In the form $y=a(x-h)^2+k,$ the given equation, $y=-0.5(x+7)^2-3
,$ is equivalent to

$$
begin{align*}
y&=-0.5(x-(-7))^2-3
.end{align*}
$$

Since the vertex of the quadratic equation in the form $y=a(x-h)^2+k$ is given by $(h,k),$ then the vertex is $left(
-7,-3
right).$

Since the axis of symmetry of a quadratic equation is given by $x=h,$ then the axis of symmetry is

$$
begin{align*}
x=-7
.end{align*}
$$

Since in the given equation, $a<0$ ($a=
-0.5
,$) then the parabola defined by the given equation opens
down.

d) In the form $y=a(x-h)^2+k,$ the given equation, $y=-3(x+2)(x-5)
,$ is equivalent to

$$
begin{align*}
y&=-3(x+2)(x-5)
\
y&=-3(x^2-3x-10)
\
y&=-3(x^2-3x)+30
\
y&=-3left( x^2-3x+dfrac{9}{4} right)+30+3left( dfrac{9}{4} right)
\
y&=-3left( x-dfrac{3}{2} right)^2+30+dfrac{27}{4}
\
y&=-3left( x-dfrac{3}{2} right)^2+dfrac{120}{4}+dfrac{27}{4}
\
y&=-3left( x-dfrac{3}{2} right)^2+dfrac{147}{4}
.end{align*}
$$

Since the vertex of the quadratic equation in the form $y=a(x-h)^2+k$ is given by $(h,k),$ then the vertex is $left(
dfrac{3}{2}, dfrac{147}{4}
right).$

Since the axis of symmetry of the quadratic equation is given by $x=h,$ then the axis of symmetry is

$$
begin{align*}
x=dfrac{3}{2}
.end{align*}
$$

Since in the given equation, $a<0$ ($a=
-3
,$) then the parabola defined by the given equation opens
down.

a) $text{Vertex: }left(
0,4
right);$ $text{Axis of Symmetry: }
x=0
;$ $text{Opening: }text{
Up
}$

b) $text{Vertex: }left(
4,1
right);$ $text{Axis of Symmetry: }
x=4
;$ $text{Opening: }text{
Up
}$

c) $text{Vertex: }left(
-7,-3
right);$ $text{Axis of Symmetry: }
x=-7
;$ $text{Opening: }text{
Down
}$

d) $text{Vertex: }left(
dfrac{3}{2}, dfrac{147}{4}
right);$ $text{Axis of Symmetry: }
x=dfrac{3}{2}
;$ $text{Opening: }text{
Down
}$

In solving quadratic equations $ax^2+bx+c=0$, we can use factoring (if factorable) or quadratic formula (for general case)

$$
begin{equation*} x=dfrac{-bpmsqrt{b^2-4ac}}{2a} end{equation*}
$$

Notice that if $b^2-4ac<0$, there will be no real solution.

$bold{a);;}$ We shall try to find factors of 24 whose sum is $-11$ and we observe that it must $-8$ and $-3$, thus

$$
begin{align*} x^2-11x+24&=(x-8)(x-3)\
& (x-8)=0;text{or};(x-3)=0\
therefore; & x=8;text{or} ;;x=3end{align*}
$$

$bold{b);;}$ We cannot find factors of 3 whose sum is $-6$, so we shall use quadratic formula.

$a=1$ , $b=-6$ , $c=3$

$$
begin{align*} x&=dfrac{-bpmsqrt{b^2-4ac}}{2a}\
&=dfrac{6pmsqrt{(-6)^2-4(1)(3)}}{2(1)}\
&=dfrac{6pmsqrt{24}}{2}\
&=dfrac{6pmsqrt{4times 6}}{2}\
&=dfrac{6pm2sqrt{6}}{2}\
x&=3pmsqrt{6}\
therefore; &; x=0.55;text{or} ;5.45end{align*}
$$

$bold{c);;}$ This is factorable. We shall list factors of 3 and $-5$ that would fit the middle term. By trial and error, we get

$$
begin{align*} 3x^2-2x-5&=(3x-5)(x+1) \
&(3x-5)=0 ;text{ or } x+1=0\
therefore; &x=dfrac{5}{3};text{or};-1\
&x=1.67;text{or} -1 end{align*}
$$

$bold{d);;}$ We shall rewrite this in the form $ax^2+bx+c=0$

$(3x^2-x^2)+(2x-9x)+3=0implies 2x^2-7x+3=0$

$$
begin{align*} 2x^2-7x+3&=(2x-1)(x-3)\
&(2x-1)=0;text{ or }; (x-3)=0\
therefore; &x=frac{1}{2};text{ or };x=3\
end{align*}
$$

a) $x=$ 3 or 8

b) $x=0.55$ or 5.45

c) $x=-1$ or 1.67

d) $x=0.5$ or 3

The $x$-intercept of each function are the values of $x$ when $f(x)=0$.

We can find it either by factoring or quadratic formula.

$ax^2+bx+c=0$

$$
x=dfrac{-bpmsqrt{b^2-4ac}}{2a}
$$

a) This is difference of two squares $a^2-b^2=(a-b)(a+b)$

$f(x)=x^2-9=(x-3)(x+3)=0$

$x=3$ or $x=-3$

b) We cannot factor this so we’ll use quadratic formula.

$a=1$ , $b=-8$ , $c=-18$

$x=dfrac{-(-8)pmsqrt{(-8)^2-4(1)(-18)}}{2(1)}$

$x=dfrac{8pm sqrt{136}}{2}=dfrac{8pmsqrt{4cdot 34}}{2}$

$x=dfrac{8pm 2sqrt{34}}{2}$

$x=4pm sqrt{34}$

$x=-1.83$ or $x=9.83$

c) We can factor this by trial and error

$f(x)=-3x^2+10x-8=(-1)(3x-4)(x-2)$

$3x-4=0implies x=dfrac{4}{3}=1.67$

$$
x-2=0implies x=2
$$

d) We can factor out $x$

$f(x)=6x-x^2$

$f(x)=x(6-x)$

$x=0$ or $x=6$

a) $x=3$ or 8

b) $x=-1.83$ or $9.83$

c) $x=1.67$ or 2

d) $x=0$ or $6$

Definition: A quadratic function is a function of the form $f(x)=ax^2+bx+c.$

Characteristics: The graph of a quadratic function is a parabola that opens up or down.

Examples:

$$
f(x)=2x^2+3x-1
$$

$$
g(x)=-3(x-2)^2+1
$$

Non-examples:

$$
f(x)=x+1
$$

$$
g(x)=x^3-1
$$