Question

NO LINKS!!! What is the equation of these two graphs?​

Answer 1

Answer:

$\textsf{1.} \quad y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}\:\:\textsf{ where}\:\:x \geq \dfrac{15-\sqrt{769}}{2}\\\\\quad \textsf{or} \quad y=2\sqrt{x+5}$

$\textsf{2.} \quad y=-|x+1|+5$

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:

$y=ax^2+bx+c$

Given points:

• (-4, 2)
• (-1, 4)
• (4, 6)

Substitute the given points into the equation to create 3 equations:

Equation 1  (-4, 2)

$\implies a(-4)^2+b(-4)+c=2$

$\implies 16a-4b+c=2$

Equation 2  (-1, 4)

$\implies a(-1)^2+b(-1)+c=4$

$\implies a-b+c=4$

Equation 3  (4, 6)

$\implies a(4)^2+b(4)+c=6$

$\implies 16a+4b+c=6$

Subtract Equation 1 from Equation 3 to eliminate variables a and c:

$\implies (16a+4b+c)-(16a-4b+c)=6-2$

$\implies 8b=4$

$\implies b=\dfrac{4}{8}$

$\implies b=\dfrac{1}{2}$

Subtract Equation 2 from Equation 3 to eliminate variable c:

$\implies (16a+4b+c)-(a-b+c)=6-4$

$\implies 15a+5b=2$

$\implies 15a=2-5b$

$\implies a=\dfrac{2-5b}{15}$

Substitute found value of b into the expression for a and solve for a:

$\implies a=\dfrac{2-5(\frac{1}{2})}{15}$

$\implies a=-\dfrac{1}{30}$

Substitute found values of a and b into Equation 2 and solve for c:

$\implies a-b+c=4$

$\implies -\dfrac{1}{30}-\dfrac{1}{2}+c=4$

$\implies c=\dfrac{68}{15}$

Therefore, the equation of the graph is:

$y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}$

$\textsf{with the restricted domain}: \quad x \geq \dfrac{15-\sqrt{769}}{2}$

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:

$y=\sqrt{x}$

Translated 5 units left so that the x-intercept is (0, -5):

$\implies y=\sqrt{x+5}$

The curve is stretched vertically, so:

$\implies y=a\sqrt{x+5} \quad \textsf{(where a is some constant)}$

To find a, substitute the coordinates of the given points:

$\implies a\sqrt{-4+5}=2$

$\implies a=2$

$\implies a\sqrt{-1+5}=4$

$\implies 2a=4$

$\implies a=2$

$\implies a\sqrt{4+5}=6$

$\implies 3a=6$

$\implies a=2$

As the value of a is the same for all points, the equation of the line is:

$y=2\sqrt{x+5}$

Question 2

Vertex form of an absolute value function

$f(x)=a|x-h|+k$

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:

$\implies a|0-(-1)|+5=4$

$\implies a+5=4$

$\implies a=-1$

Substituting the given vertex and the found value of a into the function, the equation of the graph is:

$y=-|x+1|+5$