Graph Quadratic Functions in Standard Form 2

Question 1 of 4

Graph

y = - x^{2} + 6 x - 9

$y=-x^2+6x-9$ .

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Standard Form of a Parabola

y = a x^{2} + b x + c

$\color{green}{y}=a \color{green}{x}^{2}+b\color{green}{x}+c$

The value of

a

$a$ is negative, so the graph is concave down.

Solve for the

y

$y$ -intercept by substituting $x = 0$ $x=0$ .

$y$ $y$	$=$ $=$	$-\color{green}{x}^{2}+6\color {green}{x}-9$
	$=$ $=$	$-(\color{green}{0}^{2})-6(\color {green}{0})-9$	Substitute $x = 0$ $x=0$
	$=$ $=$	$0 - 0 - 9$ $0-0-9$
$y$ $y$	$=$ $=$	$- 9$ $-9$	Simplify

Find the vertex of the parabola by solving for

x

$x$ from the formula

x = - \frac{b}{2 a}

$x=-b/(2a)$ .

$x$ $x$	$=$ $=$	$- \frac{b}{2 a}$ $-b/(2a)$

	$=$ $=$	$- \frac{6}{2 (- 1)}$ $-6/(2(-1))$	$a = - 1$ $a=-1$ , $b = 6$ $b=6$

	$=$ $=$	$- \frac{6}{- 2}$ $-6/(-2)$

$x$ $x$	$=$ $=$	$3$ $3$

This also marks the axis of symmetry.

Substitute the value of

x

$x$ back into the quadratic equation to find the vertex.

$y$ $y$	$=$ $=$	$-\color{green}{x}^{2}+6\color {green}{x}-9$
	$=$ $=$	$-(\color{green}{3}^{2})+6(\color {green}{3})-9$	Substitute $x = 3$ $x=3$
	$=$ $=$	$- 9 + 18 - 9$ $-9+18-9$
$y$ $y$	$=$ $=$	$0$ $0$	Simplify

This corresponds to a vertex of

(3, 0)

$(3,0)$ . Mark this on the graph.

Using the vertex and the axis of symmetry obtained above, a graph can now be drawn.

Question 2 of 4

Graph the function

y = 2 x^{2} - 5 x - 3

$y=2x^2-5x-3$

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Axis of Symmetry

x = \frac{- b}{2 a}

$x=\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$

First, find the

x

$x$ values by factoring using cross method

(2 x + 1) (x - 3)

$(2x+1)(x-3)$

=

$=$

0

$0$

$2 x + 1$ $2x+1$	$=$ $=$	$0$ $0$
$2 x + 1$ $2x+1$ $- 1$ $-1$	$=$ $=$	$0$ $0$ $- 1$ $-1$
$2 x$ $2x$	$=$ $=$	$- 1$ $-1$
$2 x$ $2x$ $\div 2$ $-:2$	$=$ $=$	$- 1$ $-1$ $\div 2$ $-:2$
$x$ $x$	$=$ $=$	$- \frac{1}{2}$ $-1/2$

$x - 3$ $x-3$	$=$ $=$	$0$ $0$
$x - 3$ $x-3$ $+ 3$ $+3$	$=$ $=$	$0$ $0$ $+ 3$ $+3$
$x$ $x$	$=$ $=$	$3$ $3$

Mark these

2

$2$ points on the

x

$x$ axis

Next, find the axis of symmetry and add it to the graph

y = 2 x^{2} - 5 x - 3

$y=2x^2-5x-3$

$a = 2$ $a=2$ $b = - 5$ $b=-5$ $c = - 3$ $c=-3$

$x$ $x$	$=$ $=$	$\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$	Axis of Symmetry

$x$ $x$	$=$ $=$	$\frac{-\color{#9a00c7}{(-5)}}{2\color{#00880A}{(2)}}$	Substitute values

$x$ $x$	$=$ $=$	$\frac{5}{4}$ $5/4$

Now, substitute

x = \frac{5}{4}

$x=5/4$ to the equation to know where the graph intersects the axis of symmetry

$y$ $y$	$=$ $=$	$2 x^{2} - 5 x - 3$ $2x^2-5x-3$

	$=$ $=$	$2 {(\frac{5}{4})}^{2} - 5 (\frac{5}{4}) - 3$ $2(5/4)^2-5(5/4)-3$	Substitute $x = \frac{5}{4}$ $x=5/4$

	$=$ $=$	$- \frac{49}{8}$ $-49/8$

$y$ $y$	$=$ $=$	$- 6 \frac{1}{8}$ $-6 1/8$

Finally, substitute

x = 0

$x=0$ to find where graph intersects the

y

$y$ axis

$y$ $y$	$=$ $=$	$2 x^{2} - 5 x - 3$ $2x^2-5x-3$
	$=$ $=$	$2 {(0)}^{2} - 5 (0) - 3$ $2(0)^2-5(0)-3$	Substitute $x = 0$ $x=0$
	$=$ $=$	$0 - 0 - 3$ $0-0-3$
$y$ $y$	$=$ $=$	$- 3$ $-3$

Simply connect the points to form a parabola

Question 3 of 4

Graph the function

y = - x^{2} + 2 x + 15

$y=-x^2+2x+15$

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Axis of Symmetry

x = \frac{- b}{2 a}

$x=\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$

First, find the

x

$x$ values by factoring using cross method

(x + 3) (- x + 5)

$(x+3)(-x+5)$

=

$=$

0

$0$

$x + 3$ $x+3$	$=$ $=$	$0$ $0$
$x + 3$ $x+3$ $- 3$ $-3$	$=$ $=$	$0$ $0$ $- 3$ $-3$
$x$ $x$	$=$ $=$	$- 3$ $-3$

$- x + 5$ $-x+5$	$=$ $=$	$0$ $0$
$- x + 5$ $-x+5$ $+ x$ $+x$	$=$ $=$	$0$ $0$ $+ x$ $+x$
$5$ $5$	$=$ $=$	$x$ $x$
$x$ $x$	$=$ $=$	$5$ $5$

Mark these

2

$2$ points on the

x

$x$ axis

Next, find the axis of symmetry and add it to the graph

y = - x^{2} + 2 x + 15

$y=-x^2+2x+15$

$a = - 1$ $a=-1$ $b = 2$ $b=2$ $c = 15$ $c=15$

$x$ $x$	$=$ $=$	$\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$	Axis of Symmetry

$x$ $x$	$=$ $=$	$\frac{-\color{#9a00c7}{2}}{2\color{#00880A}{(-1)}}$	Substitute values

$x$ $x$	$=$ $=$	$\frac{- 2}{- 2}$ $(-2)/(-2)$

$x$ $x$	$=$ $=$	$1$ $1$

Now, substitute

x = 1

$x=1$ to the equation to know where the graph intersects the axis of symmetry

$y$ $y$	$=$ $=$	$- x^{2} + 2 x + 15$ $-x^2+2x+15$
	$=$ $=$	$- {(1)}^{2} + 2 (1) + 15$ $-(1)^2+2(1)+15$	Substitute $x = 1$ $x=1$

	$=$ $=$	$- 1 + 2 + 15$ $-1+2+15$
$y$ $y$	$=$ $=$	$16$ $16$

Finally, substitute

x = 0

$x=0$ to find where graph intersects the

y

$y$ axis

$y$ $y$	$=$ $=$	$- x^{2} + 2 x + 15$ $-x^2+2x+15$
	$=$ $=$	$- 0^{2} + 2 (0) + 15$ $-0^2+2(0)+15$	Substitute $x = 0$ $x=0$
	$=$ $=$	$- 0 + 0 + 15$ $-0+0+15$
$y$ $y$	$=$ $=$	$15$ $15$

Simply connect the points to form a parabola

Question 4 of 4

Graph the function

y = - x^{2} + 3 x - 5

$y=-x^2+3x-5$

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Axis of Symmetry

x = \frac{- b}{2 a}

$x=\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$

The Quadratic Formula

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$

First, find the

x

$x$ values using the quadratic formula

y = - x^{2} + 3 x - 5

$y=-x^2+3x-5$

$a = - 1$ $a=-1$ $b = 3$ $b=3$ $c = - 5$ $c=-5$

$x$ $x$	$=$ $=$	$\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$	Quadratic Formula

	$=$ $=$	$\frac {- \color{#9a00c7}{3} \pm \sqrt {\color{#9a00c7}{3}^2-4\color{#00880A}{(-1)}\color{#007DDC}{(-5)}} }{2 \color{#00880A}{(-1)}}$	Plug in the values of $a, b$ $a, b$ and $c$ $c$

	$=$ $=$	$\frac {-3 \pm \sqrt {9 -20} }{-2}$

	$=$ $=$	$\frac {-3 \pm \sqrt {-11} }{-2}$

Remember that a negative value inside a surd gives out imaginary roots, hence

x

$x$ has no solution

This means the graph does not touch the

x

$x$ axis

Next, find the axis of symmetry

$a = - 1$ $a=-1$ $b = 3$ $b=3$ $c = - 5$ $c=-5$

$x$ $x$	$=$ $=$	$\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$	Axis of Symmetry

$x$ $x$	$=$ $=$	$\frac{-\color{#9a00c7}{3}}{2\color{#00880A}{(-1)}}$	Substitute values

$x$ $x$	$=$ $=$	$\frac{- 3}{- 2}$ $(-3)/(-2)$

$x$ $x$	$=$ $=$	$\frac{3}{2}$ $3/2$

Now, substitute

x = \frac{3}{2}

$x=3/2$ to the equation to know where the graph intersects the axis of symmetry

$y$ $y$	$=$ $=$	$- x^{2} + 3 x - 5$ $-x^2+3x-5$

	$=$ $=$	$- \frac{9}{4} + \frac{9}{2} - 5$ $-9/4+9/2-5$	Substitute $x = \frac{3}{2}$ $x=3/2$

$y$ $y$	$=$	$-2 3/4$

Finally, substitute

$x=0$ to find where graph intersects the

$y$ axis

$y$	$=$	$-x^2+3x-5$
	$=$	$-0^2+3(0)-5$	Substitute $x=0$
	$=$	$-0+0-5$
$y$	$=$	$-5$

Simply connect the points to form a parabola

Graph Quadratic Functions in Standard Form 2

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Quiz summary

Information

1. Question

Hint

Well Done!

Standard Form of a Parabola

2. Question

Hint

Keep Going!

Axis of Symmetry

3. Question

Hint

Well Done!

Axis of Symmetry

4. Question

Hint

Correct!

Axis of Symmetry

The Quadratic Formula

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