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Graph Quadratic Functions in Standard Form 2Graph Quadratic Functions in Standard Form 2
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Question 1 of 4
1. Question
Graph `y=-x^2+6x-9`.Hint
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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.Solve for the `y`-intercept by substituting `x=0`.`y` `=` $$-\color{green}{x}^{2}+6\color {green}{x}-9$$ `=` $$-(\color{green}{0}^{2})-6(\color {green}{0})-9$$ Substitute `x=0` `=` `0-0-9` `y` `=` `-9` Simplify Find the vertex of the parabola by solving for `x` from the formula `x=-b/(2a)`.`x` `=` `-b/(2a)` `=` `-6/(2(-1))` `a=-1`,`b=6` `=` `-6/(-2)` `x` `=` `3` This also marks the axis of symmetry.Substitute the value of `x` back into the quadratic equation to find the vertex.`y` `=` $$-\color{green}{x}^{2}+6\color {green}{x}-9$$ `=` $$-(\color{green}{3}^{2})+6(\color {green}{3})-9$$ Substitute `x=3` `=` `-9+18-9` `y` `=` `0` Simplify This corresponds to a vertex of `(3,0)`. Mark this on the graph.
Using the vertex and the axis of symmetry obtained above, a graph can now be drawn.
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Question 2 of 4
2. Question
Graph the function`y=2x^2-5x-3`Hint
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Axis of Symmetry
$$x=\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$$First, find the `x` values by factoring using cross method
`(2x+1)(x-3)` `=` `0` `2x+1` `=` `0` `2x+1` `-1` `=` `0` `-1` `2x` `=` `-1` `2x``-:2` `=` `-1``-:2` `x` `=` `-1/2` `x-3` `=` `0` `x-3` `+3` `=` `0` `+3` `x` `=` `3` Mark these `2` points on the `x` axis
Next, find the axis of symmetry and add it to the graph`y=2x^2-5x-3``a=2` `b=-5` `c=-3``x` `=` $$\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$$ Axis of Symmetry `x` `=` $$\frac{-\color{#9a00c7}{(-5)}}{2\color{#00880A}{(2)}}$$ Substitute values `x` `=` `5/4` 
Now, substitute `x=5/4` to the equation to know where the graph intersects the axis of symmetry`y` `=` `2x^2-5x-3` `=` `2(5/4)^2-5(5/4)-3` Substitute `x=5/4` `=` `-49/8` `y` `=` `-6 1/8` 
Finally, substitute `x=0` to find where graph intersects the `y` axis`y` `=` `2x^2-5x-3` `=` `2(0)^2-5(0)-3` Substitute `x=0` `=` `0-0-3` `y` `=` `-3` 
Simply connect the points to form a parabola
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Question 3 of 4
3. Question
Graph the function`y=-x^2+2x+15`Hint
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Axis of Symmetry
$$x=\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$$First, find the `x` values by factoring using cross method
`(x+3)(-x+5)` `=` `0` `x+3` `=` `0` `x+3``-3` `=` `0``-3` `x` `=` `-3` `-x+5` `=` `0` `-x+5``+x` `=` `0``+x` `5` `=` `x` `x` `=` `5` Mark these `2` points on the `x` axis
Next, find the axis of symmetry and add it to the graph`y=-x^2+2x+15``a=-1` `b=2` `c=15``x` `=` $$\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$$ Axis of Symmetry `x` `=` $$\frac{-\color{#9a00c7}{2}}{2\color{#00880A}{(-1)}}$$ Substitute values `x` `=` `(-2)/(-2)` `x` `=` `1` 
Now, substitute `x=1` to the equation to know where the graph intersects the axis of symmetry`y` `=` `-x^2+2x+15` `=` `-(1)^2+2(1)+15` Substitute `x=1` `=` `-1+2+15` `y` `=` `16` 
Finally, substitute `x=0` to find where graph intersects the `y` axis`y` `=` `-x^2+2x+15` `=` `-0^2+2(0)+15` Substitute `x=0` `=` `-0+0+15` `y` `=` `15` 
Simply connect the points to form a parabola
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Question 4 of 4
4. Question
Graph the function`y=-x^2+3x-5`Hint
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Axis of Symmetry
$$x=\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$$The Quadratic Formula
$$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$$First, find the `x` values using the quadratic formula`y=-x^2+3x-5``a=-1` `b=3` `c=-5``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` and `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` has no solutionThis means the graph does not touch the `x` axisNext, find the axis of symmetry`a=-1` `b=3` `c=-5``x` `=` $$\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$$ Axis of Symmetry `x` `=` $$\frac{-\color{#9a00c7}{3}}{2\color{#00880A}{(-1)}}$$ Substitute values `x` `=` `(-3)/(-2)` `x` `=` `3/2` 
Now, substitute `x=3/2` to the equation to know where the graph intersects the axis of symmetry`y` `=` `-x^2+3x-5` `=` `-9/4+9/2-5` Substitute `x=3/2` `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
Quizzes
- Sum & Product of Roots 1
- Sum & Product of Roots 2
- Sum & Product of Roots 3
- Sum & Product of Roots 4
- Solving Equations by Factoring 1
- Solving Equations Using the Quadratic Formula
- Completing the Square 1
- Completing the Square 2
- Intro to Quadratic Functions (Parabolas) 1
- Intro to Quadratic Functions (Parabolas) 2
- Intro to Quadratic Functions (Parabolas) 3
- Graph Quadratic Functions in Standard Form 1
- Graph Quadratic Functions in Standard Form 2
- Graph Quadratic Functions by Completing the Square
- Graph Quadratic Functions in Vertex Form
- Write a Quadratic Equation from the Graph
- Write a Quadratic Equation Given the Vertex and Another Point
- Quadratic Inequalities 1
- Quadratic Inequalities 2
- Quadratics Word Problems 1
- Quadratics Word Problems 2
- Quadratic Identities
- Graphing Quadratics Using the Discriminant
- Positive and Negative Definite
- Applications of the Discriminant 1
- Applications of the Discriminant 2
- Solving Reducible Equations