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Exercise 1. Une graphical method to solve the following. it. x^2-1=0 11 x^2+2x+1=0 x^2+3x-4=0 d x^2-4x+6=0 The quadratic function I'interseets the x-axis at the points (1,0) and (-4,0) What is the solution Net of the equation f(x)=0 At what values of does the graph of the equation y=(x+2)(x-6) Naxis?

Pergunta

Exercise
1. Une graphical method to solve the following.
it. x^2-1=0
11 x^2+2x+1=0
x^2+3x-4=0
d x^2-4x+6=0
The quadratic function I'interseets the x-axis at the points (1,0) and (-4,0)
What is the solution Net of the equation f(x)=0
At what values of does the graph of the equation y=(x+2)(x-6)
Naxis?

Exercise 1. Une graphical method to solve the following. it. x^2-1=0 11 x^2+2x+1=0 x^2+3x-4=0 d x^2-4x+6=0 The quadratic function I'interseets the x-axis at the points (1,0) and (-4,0) What is the solution Net of the equation f(x)=0 At what values of does the graph of the equation y=(x+2)(x-6) Naxis?

Solução

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To solve the given quadratic equations using the graphical method, we need to find the points where the graph of the quadratic function intersects the x-axis. These points represent the solutions to the equation.<br /><br />a) $x^{2}-1=0$<br />To find the solutions, we set the equation equal to zero and solve for x:<br />$x^{2}-1=0$<br />$x^{2}=1$<br />$x=\pm 1$<br /><br />So, the solutions to the equation $x^{2}-1=0$ are $x=1$ and $x=-1$.<br /><br />b) $x^{2}+2x+1=0$<br />To find the solutions, we set the equation equal to zero and solve for x:<br />$x^{2}+2x+1=0$<br />$(x+1)^{2}=0$<br />$x+1=0$<br />$x=-1$<br /><br />So, the solution to the equation $x^{2}+2x+1=0$ is $x=-1$.<br /><br />c) $x^{2}+3x-4=0$<br />To find the solutions, we set the equation equal to zero and solve for x:<br />$x^{2}+3x-4=0$<br />$(x+4)(x-1)=0$<br />$x+4=0$ or $x-1=0$<br />$x=-4$ or $x=1$<br /><br />So, the solutions to the equation $x^{2}+3x-4=0$ are $x=-4$ and $x=1$.<br /><br />d) $x^{2}-4x+6=0$<br />To find the solutions, we set the equation equal to zero and solve for x:<br />$x^{2}-4x+6=0$<br />This quadratic equation does not have real solutions as the discriminant is negative.<br /><br />So, the equation $x^{2}-4x+6=0$ has no real solutions.<br /><br />The quadratic function intersects the x-axis at the points $(1,0)$ and $(-4,0)$. Therefore, the solutions to the equation $f(x)=0$ are $x=1$ and $x=-4$.<br /><br />At what values of x does the graph of the equation $y=(x+2)(x-6)$ intersect the x-axis?<br />To find the values of x where the graph intersects the x-axis, we set the equation equal to zero and solve for x:<br />$(x+2)(x-6)=0$<br />$x+2=0$ or $x-6=0$<br />$x=-2$ or $x=6$<br /><br />So, the graph of the equation $y=(x+2)(x-6)$ intersects the x-axis at the points $x=-2$ and $x=6$.
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