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Question: How do you find the x-intercept of the graph \[y=4{{x}^{2}}+11x-3\] ?...

How do you find the x-intercept of the graph y=4x2+11x3y=4{{x}^{2}}+11x-3 ?

Explanation

Solution

In order to solve the solve question for x-intercept, we need to put the value of y = 0 and solve for the value of ‘x’. By putting y = 0 we will then get a quadratic equation. The general form of quadratic equation isax2+bx+c=0a{{x}^{2}}+bx+c=0, where a, b and c are the numerical coefficients or constants, and the value of xxis unknown one fundamental rule is that the value of a, the first constant can never be zero. Using the quadratic formula we will find the values of ‘x’.

Complete step by step answer:
We have given that,
y=4x2+11x3y=4{{x}^{2}}+11x-3
Now,
Finding the x-intercept,
We need to put the value of y = 0,
We have,
y=4x2+11x3y=4{{x}^{2}}+11x-3
0=4x2+11x3\Rightarrow 0=4{{x}^{2}}+11x-3
Rewrite the above equation as,
4x2+11x3=04{{x}^{2}}+11x-3=0
The quadratic formula provides the solution for the quadratic equation:
ax2+bx+c=0a{{x}^{2}}+bx+c=0
In which a, b and c are the coefficient of respectively terms in the quadratic equation, as follows:
Roots of the quadratic equation= b±b24ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Determine the quadratic equation’s coefficients a, b and c:
The coefficient of the given quadratic equation 4x2+11x3=04{{x}^{2}}+11x-3=0are,
a=4 b=11 c=3a = 4\\\ \Rightarrow b = 11\\\ \Rightarrow c = -3
Plug these coefficient into the quadratic formula:
b±b24ac2a=11±(11)2(4×4×3)2×4\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-11\pm \sqrt{{{\left( 11 \right)}^{2}}-\left( 4\times 4\times -3 \right)}}{2\times 4}
Solve exponents and square root, we get
11±(11)2(4×4×3)2×4\Rightarrow \dfrac{-11\pm \sqrt{{{\left( 11 \right)}^{2}}-\left( 4\times 4\times -3 \right)}}{2\times 4}
Performing any multiplication and division given in the formula,
11±121+488\Rightarrow \dfrac{-11\pm \sqrt{121+48}}{8}
11±1698=11±138\Rightarrow \dfrac{-11\pm \sqrt{169}}{8}=\dfrac{-11\pm 13}{8}
We got two values, i.e.
11+138 and 11138\Rightarrow \dfrac{-11+13}{8}\ and\ \dfrac{-11-13}{8}
Solving the above, we get
28 and 248\Rightarrow \dfrac{2}{8}\ and\ \dfrac{-24}{8}
Converting into simplest form,
14 and 3\Rightarrow \dfrac{1}{4}\ and\ -3
Therefore,
x=14,3\therefore x=\dfrac{1}{4},-3

Therefore,the possible values of x-intercept are 3 and 14-3\ and\ \dfrac{1}{4}.

Note: While solving these types of questions, students need to know the concept of finding intercept. Solve the equation very carefully and do the calculation part very explicitly to avoid making any errors. They should be well aware about the concept of finding the intercept when given parabola, quadratic equation, vertex form etc.