Chapter Review
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Make sure the equation is written in standard form. Identify the related function and graph it.
x=- 4
We are asked to solve the given quadratic equation by graphing. There are three steps to solving a quadratic equation by graphing.
The solutions of ax^2+bx+c=0 are the x-intercepts of the graph of y=ax^2+bx+c. Let's write our equation in standard form. This means gathering all of the terms on the left-hand side of the equation.
LHS-x^2=RHS-x^2
Commutative Property of Addition
Now we can identify the function related to the equation. Equation:&- x^2-8x-16=0 Related Function:&y=- x^2-8x-16
To draw the graph of the related function written in standard form, we must start by identifying the values of a, b, and c. y=- x^2-8x-16 ⇔ y= - 1x^2+( - 8)x+( - 16) We can see that a= - 1, b= - 8, and c= - 16. Now, we will follow four steps to graph the function.
The axis of symmetry is a vertical line with equation x=- b2 a. Since we already know the values of a and b, we can substitute them into the formula.
The axis of symmetry of the parabola is the vertical line with equation x=- 4.
To calculate the vertex, we need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: ( - b/2 a, f( - b/2 a ) ) Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=- 4. Thus, the x-coordinate of the vertex is also - 4. To find the y-coordinate, we need to substitute - 4 for x in the given equation.
x= - 4
(- a)^2=a^2
- a(- b)=a* b
Add and subtract terms
We found the y-coordinate, and now we know that the vertex is (- 4,0).
The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Thus, the point where our graph intercepts the y-axis is (0, - 16). Let's plot this point and its reflection across the axis of symmetry.
We can now draw the graph of the function. Since a= - 1, which is negative, the parabola will open downwards. Let's connect the three points with a smooth curve.
Let's identify the x-intercepts of the graph of the related function.
We can see that the parabola intercepts the x-axis once. The point of intersection is ( - 4,0), it is also the vertex of the parabola. The equation - 8x-16=x^2 has one solution, x= - 4.
x= - 4
- a(- b)=a* b
Subtract term
(- a)^2=a^2
Since 16=16, we know that x=- 4 is a solution of the equation.