Graph Quadratic Functions in Standard Form 1

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Question 1 of 4

Graph

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

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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

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

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

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

Mark the

$y$ -intercept on the graph.

Next, solve for the

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

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

Mark the

$x$ -intercepts on the graph.

Draw a parabola using the points.

Question 2 of 4

Graph

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

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Standard Form of a Parabola

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

Rewrite the equation so that it is in standard form.

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

The value of

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

Find the

$y$ -intercept of the equation by substituting

$x=0$ .

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

Next, solve for the

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

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

Mark the

$x$ -intercepts on the graph.

Find the vertex from the formula

$x=-b/(2a)$

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

	$=$	$-8/(2(-2))$	$a=-2$ , $b=8$

	$=$	$-8/(-4)$

$x$	$=$	$2$

Solve for the

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

$x$ .

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

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

$(2,8)$

Question 3 of 4

Graph

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

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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

$a$ is positive, so the graph is concave up.

Find the vertex of the parabola by first solving for

$x$ from the formula

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

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

	$=$	$-(-4)/(2(2))$	$a=2$ , $b=-4$

	$=$	$4/4$

$x$	$=$	$1$

Substitute the value of

$x$ into the quadratic equation.

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

This corresponds to a vertex of

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

Find the

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

$c$ , which is equal to

$3$ . 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 $y=-x^2+4x+5$ .
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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 $a$ is negative, so the graph is concave down.

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

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

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

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

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

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

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

Read across to get the factors.

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

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

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

$x$ $=$ $5$

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

$x$ $=$ $-1$

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

Graph Quadratic Functions in Standard Form 1

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Quiz summary

Information

1. Question

Hint

Keep Going!

Standard Form of a Parabola

2. Question

Hint

Correct!

Standard Form of a Parabola

3. Question

Hint

Great Work!

Standard Form of a Parabola

4. Question

Hint

Nice Job!

Standard Form of a Parabola

Quizzes

$-x+5$	$=$	$0$	Solve for $x$
$x$	$=$	$5$
$x+1$	$=$	$0$	Solve for $x$
$x$	$=$	$-1$