Graph Quadratic Functions in Standard Form 1

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Question 1 of 4

1. Question

Graph

y = 3 x^{2} + 6 x

$y=3x^2+6x$ .

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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 positive, so the parabola is concave up. Solve for the

y

$y$ -intercept by substituting $x = 0$ $x=0$ into the equation.

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

Mark the

y

$y$ -intercept on the graph.

Next, solve for the

x

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

$y$ $y$	$=$ $=$	$3\color{green}{x}^{2}+6\color {green}{x}$
$0$ $0$	$=$ $=$	$3\color{green}{x}^{2}+6\color {green}{x}$	Substitute $y = 0$ $y=0$
$0$ $0$	$=$ $=$	$3 x (x + 2)$ $3x(x+2)$	Factor out $3$ $3$
$3 x (x + 2)$ $3x(x+2)$	$=$ $=$	$0$ $0$
$3 x$ $3x$	$=$ $=$	$0$ $0$	Equate factors to $0$ $0$
$x$ $x$	$=$ $=$	$0$ $0$
$x + 2$ $x+2$	$=$ $=$	$0$ $0$	Equate factors to $0$ $0$
$x$ $x$	$=$ $=$	$- 2$ $-2$

Mark the

x

$x$ -intercepts on the graph.

Draw a parabola using the points.

Question 2 of 4

2. Question

Graph

y = 8 x - 2 x^{2}

$y=8x-2x^2$ .

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Standard Form of a Parabola

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

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

Rewrite the equation so that it is in standard form.

$y$ $y$	$=$ $=$	$8 x - 2 x^{2}$ $8x-2x^2$
$y$ $y$	$=$ $=$	$- 2 x^{2} + 8 x$ $-2x^2+8x$

The value of

a

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

Find the

y

$y$ -intercept of the equation by substituting

x = 0

$x=0$ .

$y$ $y$	$=$ $=$	$-2\color{green}{x}^{2}+8\color {green}{x}$
	$=$ $=$	$-2 ( \color{green}{0}^{2})+8(\color {green}{0})$	Substitute $x = 0$ $x=0$
$y$ $y$	$=$ $=$	$0$ $0$

Next, solve for the

x

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

$y$ $y$	$=$ $=$	$-2\color{green}{x}^{2}+8\color {green}{x}$
$0$ $0$	$=$ $=$	$-2\color{green}{x}^{2}+8\color {green}{x}$	Substitute $y = 0$ $y=0$
$0$ $0$	$=$ $=$	$- 2 x (x - 4)$ $-2x(x-4)$	Factor out $- 2 x$ $-2x$
$- 2 x (x - 4)$ $-2x(x-4)$	$=$ $=$	$0$ $0$
$- 2 x$ $-2x$	$=$ $=$	$0$ $0$	Equate factors to $0$ $0$
$x$ $x$	$=$ $=$	$0$ $0$
$x - 4$ $x-4$	$=$ $=$	$0$ $0$	Equate factors to $0$ $0$
$x$ $x$	$=$ $=$	$4$ $4$

Mark the

x

$x$ -intercepts on the graph.

Find the vertex from the formula

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

$x=-b/(2a)$

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

	$=$ $=$	$- \frac{8}{2 (- 2)}$ $-8/(2(-2))$	$a = - 2$ $a=-2$ , $b = 8$ $b=8$

	$=$ $=$	$- \frac{8}{- 4}$ $-8/(-4)$

$x$ $x$	$=$ $=$	$2$ $2$

Solve for the

y

$y$ -intercept of the vertex using the obtained value of

x

$x$ .

$y$ $y$	$=$ $=$	$- 2 x^{2} + 8 x$ $-2x^2+8x$
	$=$ $=$	$- 2 (2^{2}) + 8 (2)$ $-2(2^2)+8(2)$
	$=$ $=$	$- 2 (4) + 16$ $-2(4)+16$
	$=$ $=$	$- 8 + 16$ $-8+16$
$y$ $y$	$=$ $=$	$8$ $8$

Draw a parabola using the points, together with the obtained vertex

(2, 8)

$(2,8)$

Question 3 of 4

3. Question

Graph

y = 2 x^{2} - 4 x + 3

$y=2x^2-4x+3$ .

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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 positive, so the graph is concave up.

Find the vertex of the parabola by first solving for

x

$x$ from the formula

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

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

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

	$=$ $=$	$- \frac{- 4}{2 (2)}$ $-(-4)/(2(2))$	$a = 2$ $a=2$ , $b = - 4$ $b=-4$

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

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

Substitute the value of

x

$x$ into the quadratic equation.

$y$ $y$	$=$ $=$	$2\color{green}{x}^{2}-4\color {green}{x}+3$
	$=$ $=$	$2(\color{green}{1}^{2})-4(\color {green}{1})+3$	Substitute $x = 1$ $x=1$
	$=$ $=$	$2 (1) - 4 + 3$ $2(1)-4+3$
	$=$ $=$	$2 - 1$ $2-1$
$y$ $y$	$=$ $=$	$1$ $1$	Simplify

This corresponds to a vertex of

(1, 1)

$(1,1)$ . Mark this on the graph.

Find the

y

$y$ -intercept of the graph. This is equal to

c

$c$ , which is equal to

3

$3$ . Plot this on the graph as well.

Connect the points and take note that the parabola opens up.

Question 4 of 4

4. Question
Graph $y = - x^{2} + 4 x + 5$ $y=-x^2+4x+5$ .
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- 4.
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Standard Form of a Parabola

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

The value of $a$ $a$ is negative, so the graph is concave down.

Find the $y$ $y$ -intercept of the graph. This is equal to $c$ $c$ , which is equal to $5$ $5$ . Plot this on the graph.

Find the $x$ $x$ -intercept of the graph by substituting $y = 0$ $y=0$ .

$y$ $y$ $=$ $=$ $- x^{2} + 4 x + 5$ $-x^2+4x+5$

$0$ $0$ $=$ $=$ $- x^{2} + 4 x + 5$ $-x^2+4x+5$ $y = 0$ $y=0$

Since the equation is in standard form $($ $($ $a$ $a$ $x^{2} +$ $x^2+$ $b$ $b$ $x +$ $x+$ $c$ $c$ $= 0)$ $=0)$ we can factorise using the cross method.

$-$ $-$ $x^{2} +$ $x^2+$ $4$ $4$ $x +$ $x+$ $5$ $5$ $= 0$ $=0$

To factorise, we need to find two numbers that add to $4$ $4$ and multiply to $5$ $5$

Read across to get the factors.

$(- x + 5) (x + 1)$ $(-x+5)(x+1)$

Solve for the $x$ $x$ -intercepts and plot them on the graph.

$- x + 5$ $-x+5$ $=$ $=$ $0$ $0$ Solve for $x$ $x$

$x$ $x$ $=$ $=$ $5$ $5$

$x + 1$ $x+1$ $=$ $=$ $0$ $0$ Solve for $x$ $x$

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

Connect the points and take note that the parabola opens up.