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Here are a few recommended readings before getting started with this lesson.
A quadratic equation is a polynomial equation of second degree, meaning the highest exponent of its monomials is $2.$ A quadratic equation can be written in standard form as follows.
$ax_{2}+bx+c=0$
$f(x)=2x_{2}−2x−12$  

Direction  Vertex  Axis of Symmetry  $y$intercept 
$a>0⇒$ upward  $(21 ,225 )$  $x=21 $  $(0,12)$ 
In addition to these points, the reflection of the $y$intercept across the axis of symmetry is at $(1,12),$ which is also on the parabola. Now, the graph can be drawn.
Now, find all the points on the graph that have a $y$coordinate equal to $0.$
The $x$coordinates of the identified points solve the equation $f(x)=0.$
In the example, the $x$coordinates of the points where the curve intercepts the $x$axis are $2$ and $3.$ Note that solving an equation graphically does not necessarily lead to an exact answer. To verify a solution, substitute it into $f(x)$ and evaluate the expression. Verifying the solution $x=2$ is done by evaluating $f(2).$$x=2$
Calculate power
Multiply
$a(b)=a⋅b$
$a−(b)=a+b$
Add and subtract terms
Solution  Substitute  Evaluate 

$x=2$  $f(2)=2(2)_{2}−2(2)−12$  $f(2)=0✓$ 
$x=3$  $f(3)=2(3)_{2}−2(3)−12$  $f(3)=0✓$ 
For each quadratic equation, draw its related quadratic function to determine the number of solutions.
Paulina models the flight of a rocket she built using a quadratic function, where $h$ is the height of the rocket in meters after $t$ seconds.
$LHS−192=RHS−192$
Rearrange equation
$f(t)=16t_{2}+112t−192$  

Direction  Vertex  Axis of Symmetry  $y$intercept 
$a<0⇓downward $

$(3.5,4)$  $x=3.5$  $(0,192)$ 
In addition to the vertex and the $y$intercept, the reflection of the $y$intercept across the axis of symmetry, which is at $(7,192),$ is also on the parabola. With this information, the graph of the function can be drawn.
The parabola intersects the $x$axis twice. The points of intersection are $(3,0)$ and $(4,0).$ Therefore, the equation $16t_{2}+112t−192=0$ has two solutions, $t_{1}=3$ and $t_{2}=4.$$LHS−196=RHS−196$
Rearrange equation
$g(t)=16t_{2}+112t−196$  

Direction  Vertex  Axis of Symmetry  $y$intercept 
$a<0⇓downward $

$(3.5,0)$  $x=3.5$  $(0,196)$ 
The reflection of the $y$intercept across the axis of symmetry is at $(7,196),$ which is also on the parabola. Now, the graph can be drawn.
This time, there is only one $x$intercept.
The point of intersection is $(3.5,0).$ Therefore, the equation $16t_{2}+112t−196=0$ has one solution, $t=3.5.$Tadeo wants to form a square and a right triangle whose areas are equal. The side lengths of the geometric shapes are expressed in the diagram.
The area of the square is the square of $0.5x,$ and the area of the triangle is half the product of $2$ and $0.5x+2.$
Area of Square  Area of Triangle 

$(0.5x)_{2}$  $21 (2)(0.5x+2)$ 
$0.25x_{2}$  $0.5x+2$ 
$LHS−0.5x=RHS−0.5x$
$LHS−2=RHS−2$
$y=0.25x_{2}−0.5x−2$  

Direction  Vertex  Axis of Symmetry  $y$intercept 
$a>0⇓upward $

$(1,2.25)$  $x=1$  $(0,2)$ 
The reflection of the $y$intercept across the axis of symmetry $(2,2)$ is also on the parabola. With this information, the graph can be drawn.
From the graph, the $x$intercepts of the function can be identified.
The parabola intersects the $x$axis twice. The points of intersection are $(2,0)$ and $(4,0).$ Therefore, the quadratic equation has two solutions, $x_{1}=2$ and $x_{2}=4.$Area of the Square  Area of the Triangle 

$f(x)=(0.5x)_{2}$  $g(x)=21 ⋅2⋅(0.5x+2)$ 
$f(x)=0.25x_{2}$  $g(x)=0.5x+2$ 
To draw the graphs of these functions, a table of values will be made.
$x$  $f(x)=0.25x_{2}$  $g(x)=0.5x+2$ 

$3$  $0.25(3)_{2}=2.25$  $0.5(3)+2=0.5$ 
$0$  $0.25(0)_{2}=0$  $0.5(0)+2=2$ 
$3$  $0.25(3)_{2}=2.25$  $0.5(3)+2=3.5$ 
$(a−b)_{2}=a_{2}−2ab+b_{2}$
Distribute $0.5$
Subtract term
$LHS−4x=RHS−4x$
$LHS+0.5x_{2}=RHS+0.5x_{2}$
$y=0.7x_{2}+0.8x+8$  

Direction  Vertex  Axis of Symmetry  $y$intercept 
$a<0⇓downward $

$(74 ,35288 )$  $x=74 $  $(0,8)$ 
$x=3$
Calculate power
Multiply
Add and subtract terms
The $x$intercepts of the graph can now be identified.
The parabola intersects the $x$axis twice. From the graph only one of the intercepts is identified easily, which is $(4,0).$ The other intercept is between $3$ and $2.$ However, this intercept can be ignored because $x$ represents horizontal distance and distance cannot be negative. Therefore, in this context, only $x=4$ makes sense.$x=4$
$a−a=0$