Answer:
[tex]\textsf{1.} \quad y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}\:\:\textsf{ where}\:\:x \geq \dfrac{15-\sqrt{769}}{2}\\\\\quad \textsf{or} \quad y=2\sqrt{x+5}[/tex]
[tex]\textsf{2.} \quad y=-|x+1|+5[/tex]
Step-by-step explanation:
Question 1
Method 1 - modelling as a quadratic with restricted domain
Assuming that the points given on the graph are points that the curve passes through, the curve can be modeled as a quadratic with a limited domain. Please note that as the x-intercept has not been defined on the graph, I am not including this in this first method.
Standard form of a quadratic equation:
[tex]y=ax^2+bx+c[/tex]
Given points:
Substitute the given points into the equation to create 3 equations:
Equation 1 (-4, 2)
[tex]\implies a(-4)^2+b(-4)+c=2[/tex]
[tex]\implies 16a-4b+c=2[/tex]
Equation 2 (-1, 4)
[tex]\implies a(-1)^2+b(-1)+c=4[/tex]
[tex]\implies a-b+c=4[/tex]
Equation 3 (4, 6)
[tex]\implies a(4)^2+b(4)+c=6[/tex]
[tex]\implies 16a+4b+c=6[/tex]
Subtract Equation 1 from Equation 3 to eliminate variables a and c:
[tex]\implies (16a+4b+c)-(16a-4b+c)=6-2[/tex]
[tex]\implies 8b=4[/tex]
[tex]\implies b=\dfrac{4}{8}[/tex]
[tex]\implies b=\dfrac{1}{2}[/tex]
Subtract Equation 2 from Equation 3 to eliminate variable c:
[tex]\implies (16a+4b+c)-(a-b+c)=6-4[/tex]
[tex]\implies 15a+5b=2[/tex]
[tex]\implies 15a=2-5b[/tex]
[tex]\implies a=\dfrac{2-5b}{15}[/tex]
Substitute found value of b into the expression for a and solve for a:
[tex]\implies a=\dfrac{2-5(\frac{1}{2})}{15}[/tex]
[tex]\implies a=-\dfrac{1}{30}[/tex]
Substitute found values of a and b into Equation 2 and solve for c:
[tex]\implies a-b+c=4[/tex]
[tex]\implies -\dfrac{1}{30}-\dfrac{1}{2}+c=4[/tex]
[tex]\implies c=\dfrac{68}{15}[/tex]
Therefore, the equation of the graph is:
[tex]y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}[/tex]
[tex]\textsf{with the restricted domain}: \quad x \geq \dfrac{15-\sqrt{769}}{2}[/tex]
Method 2 - modelling as a square root function
Assuming that the points given on the graph are points that the curve passes through, and the x-intercept should be included, we can model this curve as a square root function.
Given points:
- (-4, 2)
- (-1, 4)
- (4, 6)
- (0, -5)
The parent function is:
[tex]y=\sqrt{x}[/tex]
Translated 5 units left so that the x-intercept is (0, -5):
[tex]\implies y=\sqrt{x+5}[/tex]
The curve is stretched vertically, so:
[tex]\implies y=a\sqrt{x+5} \quad \textsf{(where a is some constant)}[/tex]
To find a, substitute the coordinates of the given points:
[tex]\implies a\sqrt{-4+5}=2[/tex]
[tex]\implies a=2[/tex]
[tex]\implies a\sqrt{-1+5}=4[/tex]
[tex]\implies 2a=4[/tex]
[tex]\implies a=2[/tex]
[tex]\implies a\sqrt{4+5}=6[/tex]
[tex]\implies 3a=6[/tex]
[tex]\implies a=2[/tex]
As the value of a is the same for all points, the equation of the line is:
[tex]y=2\sqrt{x+5}[/tex]
Question 2
Vertex form of an absolute value function
[tex]f(x)=a|x-h|+k[/tex]
where:
- (h, k) is the vertex
- a is some constant
From inspection of the given graph:
- vertex = (-1, 5)
- point on graph = (0, 4)
Substitute the given values into the function and solve for a:
[tex]\implies a|0-(-1)|+5=4[/tex]
[tex]\implies a+5=4[/tex]
[tex]\implies a=-1[/tex]
Substituting the given vertex and the found value of a into the function, the equation of the graph is:
[tex]y=-|x+1|+5[/tex]
